Which substitution turns a homogeneous equation dy/dx = F(y/x) into one with separable variables?

y = vx, with dy/dx = v + xdv/dx

y = v + x, with dy/dx = dv/dx + 1

A homogeneous equation is written as dx/dy = G(x/y). Which substitution, with its derivative, separates it?

x = vy, with dx/dy = v + ydv/dy

Homogeneous and Linear Differential Equations for CUET Maths

Q: What is a homogeneous function, and a homogeneous equation?

F(x, y) is homogeneous of degree n when F(λ x, λ y) = λ^n F(x, y): every term has the same total degree. x^2 + 3xy has degree 2, √(x^2 + y^2) degree 1, and y/x degree 0. The equation dy/dx = F(x, y) is homogeneous when F has degree 0, so that it can be written as a function of y/x alone.

Q: How do you solve a homogeneous equation?

Put y = vx. Then dy/dx = v + xdv/dx, the equation becomes xdv/dx = F(v) - v, and the variables separate. Integrate, then put back v = y/x.

Q: What is a linear differential equation?

dy/dx + Py = Q, where P and Q are functions of x alone, or constants. y and its derivative appear only to the first power and are never multiplied together. With x as the unknown, the form is dx/dy + Px = Q, with P and Q functions of y.

Q: What is the integrating factor?

I.F. = e^∫ P dx. Multiplying the equation by it turns the left side into d/dx(y · I.F.). Put the equation in the standard form first: xdy/dx + 2y = x^3 must be divided by x before reading P = 2/x.

Q: How do you solve a linear equation?

y · I.F. = ∫ Q · I.F. dx + C The constant goes in after integrating the right side; the integrating factor itself needs none.

About 5 min to learn · 120 questions

Section A of CUET Maths solves differential equations by separating the variables; Section B1 adds the two other methods for first-order equations, homogeneous equations and linear equations. Both follow a fixed recipe, so a little practice makes them quick marks. Learn each rule here, from homogeneous functions and the substitution y = vx to linear equations, the integrating factor and particular solutions, with worked examples, time-saving shortcuts and the traps CUET sets, then practise on free questions in CUET's own formats, each fully solved.

Syllabus area: Section B1: Mathematics

On this page · 10 sections

01Two recipes for first-order equations
02Homogeneous equations
03Solving homogeneous equations
04Linear equations and the integrating factor
05Solving linear equations
06How CUET asks this chapter
07Exam shortcuts
08Common mistakes
09Quick revision
10Check your understanding

Two recipes for first-order equations

Not every equation separates. $\frac{dy}{dx} = \frac{x^{2} + y^{2}}{2x^{2}}$ mixes $x$ and $y$ in a way no rearranging can pull apart, and $\frac{dy}{dx} + y\tan x = \sec x$ has $y$ and $x$ added together. The first is homogeneous, the second linear, and each has its own recipe.

Homogeneous equations

What is a homogeneous function, and a homogeneous equation?

$F(x, y)$ is homogeneous of degree $n$ when $F(\lambda x, \lambda y) = \lambda^{n} F(x, y)$ : every term has the same total degree. $x^{2} + 3xy$ has degree 2, $\sqrt{x^{2} + y^{2}}$ degree 1, and $\frac{y}{x}$ degree 0. The equation $\frac{dy}{dx} = F(x, y)$ is homogeneous when $F$ has degree 0, so that it can be written as a function of $\frac{y}{x}$ alone.

Watch out

A constant term spoils homogeneity: $\frac{x + y + 1}{x - y}$ mixes degree 1 with degree 0.

Solving homogeneous equations

How do you solve a homogeneous equation?

Put $y = vx$ . Then $\frac{dy}{dx} = v + x\frac{dv}{dx}$ , the equation becomes $x\frac{dv}{dx} = F(v) - v$ , and the variables separate. Integrate, then put back $v = \frac{y}{x}$ .

Example 1 · A homogeneous equation

Solve $\frac{dy}{dx} = \frac{y}{x} + \sec \frac{y}{x}$ .

With $y = vx$ : $v + x\frac{dv}{dx} = v + \sec v$ , so $\cos v\,dv = \frac{dx}{x}$ .
$\sin v = \log\lvert x \rvert + C$ .

Answer

\sin \frac{y}{x} = \log\lvert x \rvert + C

Linear equations and the integrating factor

What is a linear differential equation?

$\frac{dy}{dx} + Py = Q$ , where $P$ and $Q$ are functions of $x$ alone, or constants. $y$ and its derivative appear only to the first power and are never multiplied together. With $x$ as the unknown, the form is $\frac{dx}{dy} + Px = Q$ , with $P$ and $Q$ functions of $y$ .

What is the integrating factor?

$\text{I.F.} = e^{\int P\,dx}$ . Multiplying the equation by it turns the left side into $\frac{d}{dx}(y \cdot \text{I.F.})$ . Put the equation in the standard form first: $x\frac{dy}{dx} + 2y = x^{3}$ must be divided by $x$ before reading $P = \frac{2}{x}$ .

Solving linear equations

How do you solve a linear equation?

y \cdot \text{I.F.} = \int Q \cdot \text{I.F.}\,dx + C

The constant goes in after integrating the right side; the integrating factor itself needs none.

Example 2 · A linear equation with trigonometric coefficients

Solve $\frac{dy}{dx} + y\tan x = \sec x$ .

$P = \tan x$ , so $\text{I.F.} = e^{\int \tan x\,dx} = e^{\log \sec x} = \sec x$ .
$y\sec x = \int \sec^{2} x\,dx = \tan x + C$ , so $y = \sin x + C\cos x$ .

Answer

y = \sin x + C\cos x

Example 3 · A particular solution

Solve $\frac{dy}{dx} - y = e^{x}$ , given $y = 1$ when $x = 0$ .

$\text{I.F.} = e^{-x}$ : $ye^{-x} = \int 1\,dx = x + C$ , so $y = (x + C)e^{x}$ .
$y(0) = C = 1$ .

Answer

y = (x + 1)e^{x}

Key idea

With constant coefficients, $\frac{dy}{dx} + ay = b$ always has the solution $y = \frac{b}{a} + Ce^{-ax}$ : a steady value $\frac{b}{a}$ plus a part that dies away.

How CUET asks this chapter

Homogeneous: which function or equation is homogeneous, its degree, the substitution, a general or particular solution.
Linear: which equation is linear, its $P$ and $Q$ , its integrating factor, its solution, or $y$ at a point from a condition.
The other form: $\frac{dx}{dy} + Px = Q$ , solved for $x$ .
Formats: assertion and reason on why the integrating factor works, and statement sets on the method.

Exam shortcuts

$e^{\log f(x)} = f(x)$ : an integrating factor is often just a function, such as $x^{2}$ , $\sec x$ or $\frac{1}{x}$ .
$\frac{dy}{dx} + ay = b$ : $y = \frac{b}{a} + Ce^{-ax}$ .
Homogeneous: $y = vx$ , $\frac{dy}{dx} = v + x\frac{dv}{dx}$ .
Check a solution by substituting it back.

Common mistakes

Reading $P$ before dividing through by the coefficient of $\frac{dy}{dx}$ .
Losing the sign of $P$ : $-y\tan x$ gives $\cos x$ , not $\sec x$ .
Forgetting to multiply the constant by $x$ after $v = \frac{y}{x}$ .
Adding a constant to the integrating factor.

Quick revision

Homogeneous

$F(\lambda x, \lambda y) = \lambda^{n} F$
$y = vx$ , then separate

Linear

$\frac{dy}{dx} + Py = Q$ ; I.F. $= e^{\int P\,dx}$
$y \cdot \text{I.F.} = \int Q \cdot \text{I.F.}\,dx + C$

Next in Section B1, quantities with a direction as well as a size: vectors.

Next · 09Quick revision

Check your understanding

120 questions, grouped by type, easy to hard. Pick an answer and check it. The worked solution opens when you ask for it.

Homogeneous equationsQ1 of 120Q1

Homogeneous equations

30 questions

Question 1

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Which of these functions is homogeneous?

x^{2} + y

Correct answerYour answer

x^{2} + x y

Correct answerYour answer

x^{3} + y^{2}

Correct answerYour answer

x + \sin y

Correct answerYour answer

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AnswerB · $x^{2} + x y$

$F$ is homogeneous of degree $n$ when $F(\lambda x, \lambda y) = \lambda^{n} F(x, y)$ .
$x^{2} + x y$ passes, with degree 2; each of the others fails.

Worth noting

$x^{2} + y$ mixes degrees 2 and 1; every term must have the same degree.

Question 2

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What is the degree of the homogeneous function $x^{3} + 3 x^{2} y$ ?

0

Correct answerYour answer

1

Correct answerYour answer

2

Correct answerYour answer

3

Correct answerYour answer

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AnswerD · $3$

Replace $x, y$ by $\lambda x, \lambda y$ and see which power of $\lambda$ comes out.
It comes out as $\lambda^{3}$ , so the degree is 3.

Worth noting

2 reads the power of one term's $x$ ; the degree counts $x$ and $y$ together, $2 + 1 = 3$ .

Question 3

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Which substitution turns a homogeneous equation $\frac{dy}{dx} = F\left(\frac{y}{x}\right)$ into one with separable variables?

y = vx

, with

\frac{dy}{dx} = v

Correct answerYour answer

y = v + x

, with

\frac{dy}{dx} = \frac{dv}{dx} + 1

Correct answerYour answer

y = vx

, with

\frac{dy}{dx} = v + x\frac{dv}{dx}

Correct answerYour answer

y = \frac{v}{x}

, with

\frac{dy}{dx} = \frac{dv}{dx}

Correct answerYour answer

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AnswerC · $y = vx$ , with $\frac{dy}{dx} = v + x\frac{dv}{dx}$

Put $y = vx$ ; by the product rule $\frac{dy}{dx} = v + x\frac{dv}{dx}$ .
Then $x\frac{dv}{dx} = F(v) - v$ , which separates.

Worth noting

$\frac{dy}{dx} = v$ treats $v$ as a constant; it varies with $x$ .

Question 4

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Which of these functions is homogeneous of degree 2?

x^{2} + y^{3}

Correct answerYour answer

x y + x

Correct answerYour answer

x^{2} + 1

Correct answerYour answer

x^{2} - 3 x y + y^{2}

Correct answerYour answer

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AnswerD · $x^{2} - 3 x y + y^{2}$

Replace $x, y$ by $\lambda x, \lambda y$ ; a homogeneous function gives back $\lambda^{n}$ times itself.
$x^{2} - 3 x y + y^{2}$ gives $\lambda^{2}$ times itself; each of the others fails.

Worth noting

$xy + x$ mixes a degree-2 term with a degree-1 term.

Question 5

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The homogeneous function $F(x, y) = x^{3} + 2x^{2}y - y^{3}$ is written as $x^{3}h(v)$ , where $v = \frac{y}{x}$ . What is $h(v)$ ?

1 + 2v - v^{3}

Correct answerYour answer

v^{3} + 2v^{2} - 1

Correct answerYour answer

1 + 2v^{2} - v^{3}

Correct answerYour answer

1 + 2v - v^{2}

Correct answerYour answer

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AnswerA · $1 + 2v - v^{3}$

Divide each term by $x^{3}$ : $\frac{2x^{2}y}{x^{3}} = 2\frac{y}{x}$ and $\frac{y^{3}}{x^{3}} = \left(\frac{y}{x}\right)^{3}$ .
So $h(v) = 1 + 2v - v^{3}$ .

Worth noting

$v^{3} + 2v^{2} - 1$ is $F$ divided by $y^{3}$ , a function of $\frac{x}{y}$ , not of $v = \frac{y}{x}$ .

Question 6

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$F(x, y)$ is homogeneous of degree 3 and $F(2, -1) = 5$ . What is $F(6, -3)$ ?

5

Correct answerYour answer

15

Correct answerYour answer

45

Correct answerYour answer

135

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AnswerD · $135$

$(6, -3) = (3 \cdot 2, 3 \cdot (-1))$ , so $\lambda = 3$ .
$F(6, -3) = 3^{3}F(2, -1) = 27 \times 5 = 135$ .

Worth noting

15 multiplies by 3 only once, as if the degree were 1.

Question 7

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$f(x, y)$ is homogeneous of degree 2 and $g(x, y)$ is homogeneous of degree 3. What is the degree of the homogeneous function $f(x, y)\,g(x, y)$ ?

1

Correct answerYour answer

3

Correct answerYour answer

5

Correct answerYour answer

6

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AnswerC · $5$

$f(\lambda x, \lambda y)\,g(\lambda x, \lambda y) = \lambda^{2}f \cdot \lambda^{3}g = \lambda^{5}fg$ .
The powers of $\lambda$ add, so the degree is 5.

Worth noting

6 multiplies the degrees; for a product the powers of $\lambda$ add.

Question 8

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For $F(x, y) = 4x^{2} - 5xy$ , what is $F(\lambda x, \lambda y)$ ?

\lambda(4x^{2} - 5xy)

Correct answerYour answer

4\lambda^{2}x^{2} - 5\lambda xy

Correct answerYour answer

4\lambda x^{2} - 5\lambda^{2}xy

Correct answerYour answer

\lambda^{2}(4x^{2} - 5xy)

Correct answerYour answer

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AnswerD · $\lambda^{2}(4x^{2} - 5xy)$

Replace $x$ by $\lambda x$ and $y$ by $\lambda y$ : $4\lambda^{2}x^{2} - 5(\lambda x)(\lambda y)$ .
Both terms carry $\lambda^{2}$ , so $F(\lambda x, \lambda y) = \lambda^{2}(4x^{2} - 5xy)$ : degree 2.

Worth noting

$\lambda(4x^{2} - 5xy)$ brings in one $\lambda$ ; each of the two letters in $x^{2}$ and in $xy$ brings its own.

Question 9

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Which of these differential equations is homogeneous?

(x^{2} + xy)\,dy = (x^{2} + y^{2})\,dx

Correct answerYour answer

\frac{dy}{dx} = \frac{x + y + 1}{x - y}

Correct answerYour answer

\frac{dy}{dx} = \frac{x^{2} + y}{x}

Correct answerYour answer

\frac{dy}{dx} = x + y^{2}

Correct answerYour answer

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AnswerA · $(x^{2} + xy)\,dy = (x^{2} + y^{2})\,dx$

Written as $\frac{dy}{dx} = F(x, y)$ , the equation is homogeneous when $F$ is homogeneous of degree 0.
$F = \frac{x^{2} + y^{2}}{x^{2} + xy}$ is a ratio of two degree-2 expressions, so degree 0. The $+1$ , $x^{2} + y$ and $y^{2}$ spoil the others.

Worth noting

The $+1$ in $\frac{x + y + 1}{x - y}$ has degree 0 while $x + y$ has degree 1.

Question 10

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Which of these functions is homogeneous of degree 0?

\frac{\sin y}{x}

Correct answerYour answer

x + y^{2}

Correct answerYour answer

y e^{x}

Correct answerYour answer

\sin{\left(\frac{y}{x} \right)}

Correct answerYour answer

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AnswerD · $\sin{\left(\frac{y}{x} \right)}$

$F$ is homogeneous of degree $n$ when $F(\lambda x, \lambda y) = \lambda^{n} F(x, y)$ .
$\sin{\left(\frac{y}{x} \right)}$ passes, with degree 0; each of the others fails.

Worth noting

$\frac{\sin y}{x}$ is not homogeneous: $\sin(\lambda y)$ is not a power of $\lambda$ times $\sin y$ .

Question 11

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What is the degree of the homogeneous function $\sqrt{x^{2} + y^{2}}$ ?

0

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

1

Correct answerYour answer

2

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AnswerC · $1$

Replace $x, y$ by $\lambda x, \lambda y$ and see which power of $\lambda$ comes out.
It comes out as $\lambda^{1}$ , so the degree is 1.

Worth noting

2 is the degree under the root; the square root halves it.

Question 12

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Which of these functions is homogeneous of degree 1?

\frac{x^{2} + y^{2}}{x + y}

Correct answerYour answer

\frac{x^{2} + y}{x + 1}

Correct answerYour answer

x y + 1

Correct answerYour answer

x^{2} - y^{3}

Correct answerYour answer

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AnswerA · $\frac{x^{2} + y^{2}}{x + y}$

$F$ is homogeneous of degree $n$ when $F(\lambda x, \lambda y) = \lambda^{n} F(x, y)$ .
$\frac{x^{2} + y^{2}}{x + y}$ passes, with degree 1; each of the others fails.

Worth noting

$xy + 1$ has a constant term of degree 0 beside $xy$ of degree 2.

Question 13

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What is the degree of the homogeneous function $\frac{x^{2} - y^{2}}{x y}$ ?

-2

Correct answerYour answer

0

Correct answerYour answer

2

Correct answerYour answer

4

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AnswerB · $0$

Replace $x, y$ by $\lambda x, \lambda y$ and see which power of $\lambda$ comes out.
It comes out as $\lambda^{0}$ , so the degree is 0.

Worth noting

2 reads only the top; the bottom has degree 2 as well, so they cancel.

Question 14

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A homogeneous equation is written as $\frac{dx}{dy} = G\left(\frac{x}{y}\right)$ . Which substitution, with its derivative, separates it?

x = vy

, with

\frac{dx}{dy} = v + y\frac{dv}{dy}

Correct answerYour answer

x = vy

, with

\frac{dx}{dy} = y\frac{dv}{dy}

Correct answerYour answer

x = vy

, with

\frac{dx}{dy} = v

Correct answerYour answer

x = v + y

, with

\frac{dx}{dy} = \frac{dv}{dy}

Correct answerYour answer

Show the explanationHide the explanation

AnswerA · $x = vy$ , with $\frac{dx}{dy} = v + y\frac{dv}{dy}$

Here $x$ is the unknown, so put $x = vy$ and differentiate with respect to $y$ .
Product rule: $\frac{dx}{dy} = v + y\frac{dv}{dy}$ ; then $y\frac{dv}{dy} = G(v) - v$ separates.

Worth noting

$y\frac{dv}{dy}$ alone drops the first term of the product rule, $v \cdot 1$ .

Question 15

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Which of these differential equations is not homogeneous?

\frac{dy}{dx} = \frac{y}{x} + \sin\frac{y}{x}

Correct answerYour answer

x\,dy - y\,dx = \sqrt{x^{2} + y^{2}}\,dx

Correct answerYour answer

\frac{dy}{dx} = \frac{x^{2} + y^{2}}{xy}

Correct answerYour answer

\frac{dy}{dx} = \frac{x + y}{x^{2}}

Correct answerYour answer

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AnswerD · $\frac{dy}{dx} = \frac{x + y}{x^{2}}$

Each equation is homogeneous when its $\frac{dy}{dx} = F(x, y)$ has $F$ of degree 0.
$\frac{x + y}{x^{2}}$ has degree $1 - 2 = -1$ ; the other three right sides have degree 0.

Worth noting

$x\,dy - y\,dx = \sqrt{x^{2} + y^{2}}\,dx$ is homogeneous: $\frac{dy}{dx} = \frac{y + \sqrt{x^{2} + y^{2}}}{x}$ has degree 0.

Question 16

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Which of these functions is homogeneous?

e^{x y}

Correct answerYour answer

x + e^{y}

Correct answerYour answer

x e^{\frac{y}{x}}

Correct answerYour answer

y + \log x

Correct answerYour answer

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AnswerC · $x e^{\frac{y}{x}}$

Replace $x, y$ by $\lambda x, \lambda y$ ; a homogeneous function gives back $\lambda^{n}$ times itself.
$x e^{\frac{y}{x}}$ gives $\lambda^{1}$ times itself; each of the others fails.

Worth noting

$e^{xy}$ becomes $e^{\lambda^{2}xy}$ , which is no power of $\lambda$ times $e^{xy}$ .

Question 17

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What is the degree of the homogeneous function $\frac{x^{3} + y^{3}}{x - y}$ ?

1

Correct answerYour answer

2

Correct answerYour answer

3

Correct answerYour answer

4

Correct answerYour answer

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AnswerB · $2$

Put $\lambda x, \lambda y$ in place of $x, y$ and factor out the power of $\lambda$ .
$\lambda^{2}$ comes out, so the degree is 2.

Worth noting

3 reads only the top; the bottom's degree 1 is taken away, $3 - 1 = 2$ .

Question 18

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For which value of $a$ is $\frac{dy}{dx} = \frac{2x + ay + a - 3}{x + y}$ a homogeneous differential equation?

-3

Correct answerYour answer

0

Correct answerYour answer

2

Correct answerYour answer

3

Correct answerYour answer

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AnswerD · $3$

The bottom has degree 1, so every term on top must have degree 1: no constant term may remain.
The constant is $a - 3$ , so $a = 3$ , giving $\frac{2x + 3y}{x + y}$ of degree 0.

Worth noting

$a = 0$ removes the $ay$ term but leaves the constant $-3$ , of degree 0.

Question 19

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Read the two statements and choose the correct option.

I. If $f$ and $g$ are homogeneous of degrees 3 and 1, then $\frac{f}{g}$ is homogeneous of degree 2.

II. If $f$ and $g$ are homogeneous of degrees 2 and 1, then $f + g$ is homogeneous of degree 3.

Only ICorrect answerYour answerOnly IICorrect answerYour answerBoth I and IICorrect answerYour answerNeither I nor IICorrect answerYour answer

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AnswerA · Only I

I: $\frac{\lambda^{3}f}{\lambda g} = \lambda^{2}\frac{f}{g}$ . True.
II: $x^{2} + x$ is such a sum, and $\lambda^{2}x^{2} + \lambda x$ is no power of $\lambda$ times it, so it is not homogeneous at all. False.

Worth noting

Degrees add for a product; a sum of terms of different degrees has no single degree.

Question 20

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$F(x, y)$ is homogeneous of degree $n$ , $F(1, 2) = 3$ and $F(3, 6) = 243$ . What is $n$ ?

3

Correct answerYour answer

4

Correct answerYour answer

5

Correct answerYour answer

81

Correct answerYour answer

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AnswerB · $4$

$(3, 6) = (3 \cdot 1, 3 \cdot 2)$ , so $F(3, 6) = 3^{n}F(1, 2)$ : $3^{n} \cdot 3 = 243$ .
$3^{n} = 81 = 3^{4}$ , so $n = 4$ .

Worth noting

5 reads $243 = 3^{5}$ but forgets the factor $F(1, 2) = 3$ .

Question 21

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$F(x, y)$ is homogeneous of degree 2, and $F(1, v) = 2 + v - v^{2}$ for every $v$ . What is $F(2, 6)$ ?

-28

Correct answerYour answer

-16

Correct answerYour answer

-8

Correct answerYour answer

-4

Correct answerYour answer

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AnswerB · $-16$

$(2, 6) = (2 \cdot 1, 2 \cdot 3)$ , so $F(2, 6) = 2^{2}F(1, 3)$ .
$F(1, 3) = 2 + 3 - 9 = -4$ , so $F(2, 6) = 4 \times (-4) = -16$ .

Worth noting

$-8$ multiplies by 2 only once; degree 2 needs the factor $2^{2} = 4$ .

Question 22

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For which value of $n$ is $(x^{n}y + y^{4})\,dx = x^{2}y^{2}\,dy$ a homogeneous differential equation?

2

Correct answerYour answer

3

Correct answerYour answer

4

Correct answerYour answer

5

Correct answerYour answer

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AnswerB · $3$

$\frac{dy}{dx} = \frac{x^{n}y + y^{4}}{x^{2}y^{2}}$ ; both $x^{2}y^{2}$ and $y^{4}$ have degree 4.
So $x^{n}y$ needs degree 4 as well: $n + 1 = 4$ , $n = 3$ .

Worth noting

4 gives $x^{n}$ alone degree 4 and forgets the factor $y$ in $x^{n}y$ .

Question 23

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Assertion (A): $\frac{dy}{dx} = x + \sqrt{xy}$ is not a homogeneous differential equation.

Reason (R): For $x, y > 0$ , $x + \sqrt{xy}$ is a homogeneous function.

Both A and R are true, and R is the correct explanation of A.Correct answerYour answerBoth A and R are true, but R is not the correct explanation of A.Correct answerYour answerA is true, but R is false.Correct answerYour answerA is false, but R is true.Correct answerYour answer

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AnswerB · Both A and R are true, but R is not the correct explanation of A.

R is true: $\lambda x + \sqrt{\lambda^{2}xy} = \lambda(x + \sqrt{xy})$ , degree 1.
A is true because the equation needs $F$ of degree 0, and 1 is not 0; R says only that $F$ is homogeneous, so it does not explain A.

Worth noting

A homogeneous right side is not enough: $\frac{dy}{dx} = F$ is homogeneous only when $F$ has degree 0.

Question 24

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The equation $(x^{2} - 2y^{2})\,dx + xy\,dy = 0$ is written as $\frac{dy}{dx} = F(v)$ , where $v = \frac{y}{x}$ . What is $F(v)$ ?

\frac{1}{v} - 2v

Correct answerYour answer

2v^{2} - 1

Correct answerYour answer

2v - \frac{1}{v}

Correct answerYour answer

2v + \frac{1}{v}

Correct answerYour answer

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AnswerC · $2v - \frac{1}{v}$

$xy\,dy = -(x^{2} - 2y^{2})\,dx$ , so $\frac{dy}{dx} = \frac{2y^{2} - x^{2}}{xy}$ .
Divide top and bottom by $x^{2}$ : $\frac{2v^{2} - 1}{v} = 2v - \frac{1}{v}$ .

Worth noting

$\frac{1}{v} - 2v$ forgets the minus sign that comes from moving the $dx$ term across.

Question 25

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Match each homogeneous function in List I with its degree in List II.

List I	List II
A. $x^{3} - 2 x y^{2}$	I. $\frac{1}{2}$
B. $\frac{x^{2} - y^{2}}{x^{2} + y^{2}}$	II. $3$
C. $\sqrt{x} + \sqrt{y}$	III. $-2$
D. $\frac{1}{x^{2} + y^{2}}$	IV. $0$

A-IV, B-II, C-I, D-IIICorrect answerYour answerA-II, B-IV, C-III, D-ICorrect answerYour answerA-II, B-IV, C-I, D-IIICorrect answerYour answerA-IV, B-II, C-III, D-ICorrect answerYour answer

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AnswerC · A-II, B-IV, C-I, D-III

A: every term has degree 3. B: top and bottom both have degree 2, so the degree is $2 - 2 = 0$ .
C: $\sqrt{\lambda x} = \lambda^{1/2}\sqrt{x}$ , degree $\frac{1}{2}$ . D: $\frac{1}{\lambda^{2}(x^{2} + y^{2})}$ , degree $-2$ .

Question 26

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What is the degree of the homogeneous function $x^{2} \sin{\left(\frac{y}{x} \right)} + x y$ ?

0

Correct answerYour answer

1

Correct answerYour answer

2

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3

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AnswerC · $2$

Put $\lambda x, \lambda y$ in place of $x, y$ and factor out the power of $\lambda$ .
$\lambda^{2}$ comes out, so the degree is 2.

Worth noting

3 counts $\sin\frac{y}{x}$ as degree 1; it has degree 0, so $x^{2}\sin\frac{y}{x}$ stays at 2.

Question 27

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The equation $3y\,e^{x/y}\,dx + (y - 3x\,e^{x/y})\,dy = 0$ is written as $\frac{dx}{dy} = G(v)$ , where $v = \frac{x}{y}$ . What is $G(v)$ ?

v - \frac{1}{3}e^{-v}

Correct answerYour answer

\frac{1}{3}e^{-v} - v

Correct answerYour answer

3v - e^{-v}

Correct answerYour answer

v - \frac{1}{3}e^{v}

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AnswerA · $v - \frac{1}{3}e^{-v}$

Move the $dy$ term across, changing its sign: $\frac{dx}{dy} = \frac{3x\,e^{x/y} - y}{3y\,e^{x/y}}$ .
Divide top and bottom by $y\,e^{x/y}$ : $\frac{3v - e^{-v}}{3} = v - \frac{1}{3}e^{-v}$ .

Worth noting

$\frac{1}{3}e^{-v} - v$ keeps the sign of the $dy$ term; moving it across the equals sign flips it.

Question 28

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For $f(x, y) = y^{2}e^{x/y} + xy$ , which of the following statements are true?

I. $f$ is homogeneous of degree 2.

II. $f(3x, 3y) = 9f(x, y)$ .

III. $f(x, y) = y^{2}g\left(\frac{x}{y}\right)$ with $g(u) = e^{u} + u$ .

IV. $\frac{dy}{dx} = f(x, y)$ is a homogeneous differential equation.

All four: I, II, III and IVCorrect answerYour answerI and II onlyCorrect answerYour answerI and III onlyCorrect answerYour answerI, II and III onlyCorrect answerYour answer

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AnswerD · I, II and III only

$f(\lambda x, \lambda y) = \lambda^{2}y^{2}e^{x/y} + \lambda^{2}xy = \lambda^{2}f$ : I holds, II is the case $\lambda = 3$ , and dividing by $y^{2}$ gives III.
IV is false: $\frac{dy}{dx} = F$ is homogeneous only when $F$ has degree 0, and $f$ has degree 2.

Worth noting

IV mixes up a homogeneous function with a homogeneous equation, which needs degree 0.

Question 29

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Putting $y = vx$ in a homogeneous equation $\frac{dy}{dx} = F(x, y)$ gives $x\frac{dv}{dx} = v^{2} - 1$ . What is $F(x, y)$ ?

\frac{y^{2} + xy - x^{2}}{x^{2}}

Correct answerYour answer

\frac{y^{2} - x^{2}}{x^{2}}

Correct answerYour answer

\frac{y^{2} - xy - x^{2}}{x^{2}}

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\frac{x^{2} + xy - y^{2}}{x^{2}}

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AnswerA · $\frac{y^{2} + xy - x^{2}}{x^{2}}$

$\frac{dy}{dx} = v + x\frac{dv}{dx} = v + v^{2} - 1$ .
With $v = \frac{y}{x}$ : $\frac{y}{x} + \frac{y^{2}}{x^{2}} - 1 = \frac{y^{2} + xy - x^{2}}{x^{2}}$ .

Worth noting

$\frac{y^{2} - x^{2}}{x^{2}}$ is $v^{2} - 1$ alone; $\frac{dy}{dx}$ is $v$ plus $x\frac{dv}{dx}$ .

Question 30

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Take $x, y > 0$ . Which of these differential equations is homogeneous?

\frac{dy}{dx} = \frac{y}{x} + \log x - \log y

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\frac{dy}{dx} = \frac{y}{x} + \log x + \log y

Correct answerYour answer

\frac{dy}{dx} = \frac{y + \log x}{x}

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\frac{dy}{dx} = \frac{y}{x}\log y

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AnswerA · $\frac{dy}{dx} = \frac{y}{x} + \log x - \log y$

$\log x - \log y = -\log\frac{y}{x}$ , so $\frac{y}{x} + \log x - \log y = v - \log v$ , a function of $v = \frac{y}{x}$ alone.
$\log x + \log y$ , $\log x$ and $\log y$ each change by a multiple of $\log\lambda$ when $x, y$ become $\lambda x, \lambda y$ , so the others fail.

Worth noting

$\log x + \log y = \log xy$ gains $2\log\lambda$ ; only a difference of the two logs depends on $\frac{y}{x}$ alone.

Solving homogeneous equations

30 questions

Question 31

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Consider the differential equation $x^{2}\frac{dy}{dx} = x^{2} + xy + y^{2}$ , and put $y = vx$ .

Written as $\frac{dy}{dx} = F$ , what is $F$ in terms of $v = \frac{y}{x}$ ?

v + v^{2}

Correct answerYour answer

1 + v^{2}

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1 + v + v^{2}

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1 + 2v + v^{2}

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AnswerC · $1 + v + v^{2}$

Divide the right side by $x^{2}$ : $1 + \frac{y}{x} + \frac{y^{2}}{x^{2}}$ .
$F = 1 + v + v^{2}$ .

Worth noting

$1 + v^{2}$ is what is left after the $v$ cancels in the next step, not $F$ itself.

Question 32

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Consider the differential equation $y\,e^{x/y}\,dx = (x\,e^{x/y} + y)\,dy$ , and put $x = vy$ .

Written as $\frac{dx}{dy} = G(v)$ , what is $G(v)$ ?

v + e^{v}

Correct answerYour answer

v + 1

Correct answerYour answer

v + e^{-v}

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\frac{e^{v}}{ve^{v} + 1}

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AnswerC · $v + e^{-v}$

$\frac{dx}{dy} = \frac{x\,e^{x/y} + y}{y\,e^{x/y}} = \frac{x}{y} + \frac{y}{y\,e^{x/y}}$ .
So $G(v) = v + e^{-v}$ .

Worth noting

$v + e^{v}$ turns $\frac{1}{e^{v}}$ into $e^{v}$ ; dividing by $e^{v}$ gives $e^{-v}$ .

Question 33

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Put $y = vx$ in $\frac{dy}{dx} = \frac{x^{2} + 2y^{2}}{xy}$ . What does $x\frac{dv}{dx}$ equal?

\frac{1 + 2v^{2}}{v}

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\frac{1 + v^{2}}{v}

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\frac{1 + 3v^{2}}{v}

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\frac{1 - v^{2}}{v}

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AnswerB · $\frac{1 + v^{2}}{v}$

The right side is $\frac{1 + 2v^{2}}{v}$ , and $\frac{dy}{dx} = v + x\frac{dv}{dx}$ .
$x\frac{dv}{dx} = \frac{1 + 2v^{2}}{v} - v = \frac{1 + v^{2}}{v}$ .

Worth noting

$\frac{1 + 2v^{2}}{v}$ is the whole of $\frac{dy}{dx}$ ; the $v$ from $v + x\frac{dv}{dx}$ must be taken away.

Question 34

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A homogeneous equation $\frac{dx}{dy} = G\left(\frac{x}{y}\right)$ is solved with $x = vy$ . Which separated form does it take?

\frac{dv}{G(v) - v} = \frac{dx}{x}

Correct answerYour answer

\frac{dv}{G(v)} = \frac{dy}{y}

Correct answerYour answer

\frac{dv}{G(v) + v} = \frac{dy}{y}

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\frac{dv}{G(v) - v} = \frac{dy}{y}

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AnswerD · $\frac{dv}{G(v) - v} = \frac{dy}{y}$

$x = vy$ gives $\frac{dx}{dy} = v + y\frac{dv}{dy}$ , so $y\frac{dv}{dy} = G(v) - v$ .
Separate: $\frac{dv}{G(v) - v} = \frac{dy}{y}$ ; $y$ is the independent variable here.

Worth noting

$\frac{dx}{x}$ belongs to $y = vx$ ; with $x = vy$ the other side is $\frac{dy}{y}$ .

Question 35

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Solving a homogeneous equation with $y = vx$ , for $x > 0$ , gives $\sqrt{1 + v^{2}} = Cx$ . What is the solution in terms of $x$ and $y$ ?

\sqrt{x^{2} + y^{2}} = Cx^{2}

Correct answerYour answer

\sqrt{x^{2} + y^{2}} = Cx

Correct answerYour answer

\sqrt{1 + y^{2}} = Cx

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x^{2} + y^{2} = Cx^{2}

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AnswerA · $\sqrt{x^{2} + y^{2}} = Cx^{2}$

$\sqrt{1 + \frac{y^{2}}{x^{2}}} = \frac{\sqrt{x^{2} + y^{2}}}{x}$ for $x > 0$ .
So $\frac{\sqrt{x^{2} + y^{2}}}{x} = Cx$ , that is $\sqrt{x^{2} + y^{2}} = Cx^{2}$ .

Worth noting

$\sqrt{x^{2} + y^{2}} = Cx$ forgets the $x$ in the denominator of $\frac{\sqrt{x^{2} + y^{2}}}{x}$ .

Question 36

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What is the general solution of $\frac{dy}{dx} = \frac{x + y}{x}$ ?

y = \log\lvert x \rvert + C

Correct answerYour answer

y = x\log\lvert x \rvert + Cx

Correct answerYour answer

y = x\log\lvert x \rvert + C

Correct answerYour answer

y = Cx - x\log\lvert x \rvert

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AnswerB · $y = x\log\lvert x \rvert + Cx$

$y = vx$ : $v + x\frac{dv}{dx} = 1 + v$ , so $\frac{dv}{dx} = \frac{1}{x}$ .
$v = \log\lvert x \rvert + C$ , so $y = x\log\lvert x \rvert + Cx$ .

Worth noting

$x\log\lvert x \rvert + C$ forgets that the constant belongs to $v$ and is multiplied by $x$ too.

Question 37

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Read the two statements and choose the correct option.

I. After $y = vx$ , a homogeneous equation becomes $x\frac{dv}{dx} = F(v) - v$ .

II. The solution of a homogeneous equation is always a straight line through the origin.

Only ICorrect answerYour answerOnly IICorrect answerYour answerBoth I and IICorrect answerYour answerNeither I nor IICorrect answerYour answer

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AnswerA · Only I

I: $v + x\frac{dv}{dx} = F(v)$ , rearranged. True.
II: $y = x\log\lvert x \rvert + Cx$ , the solution of $\frac{dy}{dx} = \frac{x + y}{x}$ , is no line. False.

Question 38

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$y$ satisfies $x\frac{dy}{dx} = x + 2y$ , and $y = 0$ when $x = 1$ . What is $y$ when $x = 3$ ?

-6

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6

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9

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12

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AnswerB · $6$

The general solution is $y = Cx^{2} - x$ ; $y(1) = 0$ gives $C = 1$ .
$y = x^{2} - x$ , so at $x = 3$ : $9 - 3 = 6$ .

Worth noting

$-6$ comes from $y = Cx^{2} + x$ , the sign slip: then $C = -1$ .

Question 39

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Assertion (A): $\frac{dy}{dx} = \frac{x^{2} + y^{2}}{x^{2}}$ is a homogeneous differential equation.

Reason (R): Every equation $\frac{dy}{dx} = F(x, y)$ with $F$ a ratio of two polynomials is homogeneous.

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AnswerC · A is true, but R is false.

A is true: $F = 1 + \left(\frac{y}{x}\right)^{2}$ , a function of $\frac{y}{x}$ , degree 0.
R is false: $\frac{x + y + 1}{x - y}$ is a ratio of polynomials but not homogeneous.

Question 40

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Consider the differential equation $x^{2}\frac{dy}{dx} = x^{2} + xy + y^{2}$ , and put $y = vx$ .

After the substitution, which separated form do you get?

\frac{dv}{1 + v + v^{2}} = \frac{dx}{x}

Correct answerYour answer

\frac{dv}{1 + v^{2}} = dx

Correct answerYour answer

(1 + v^{2})\,dv = \frac{dx}{x}

Correct answerYour answer

\frac{dv}{1 + v^{2}} = \frac{dx}{x}

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AnswerD · $\frac{dv}{1 + v^{2}} = \frac{dx}{x}$

$v + x\frac{dv}{dx} = 1 + v + v^{2}$ , so $x\frac{dv}{dx} = 1 + v^{2}$ .
Separate: $\frac{dv}{1 + v^{2}} = \frac{dx}{x}$ .

Worth noting

$1 + v + v^{2}$ in the denominator forgets to take away the $v$ from $\frac{dy}{dx} = v + x\frac{dv}{dx}$ .

Question 41

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What is the general solution of $x\frac{dy}{dx} = x + 2y$ ?

y = Cx^{2} + x

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y = Cx^{2} - x

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y = Cx - x

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y = Cx^{2}

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AnswerB · $y = Cx^{2} - x$

$y = vx$ : $v + x\frac{dv}{dx} = 1 + 2v$ , so $\frac{dv}{1 + v} = \frac{dx}{x}$ .
$\log\lvert 1 + v \rvert = \log\lvert x \rvert + c$ , so $1 + \frac{y}{x} = Cx$ : $y = Cx^{2} - x$ .

Worth noting

$y = Cx^{2} + x$ moves the 1 across without changing its sign.

Question 42

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Consider the differential equation $y\,e^{x/y}\,dx = (x\,e^{x/y} + y)\,dy$ , and put $x = vy$ .

Which is the general solution of the equation?

e^{x/y} = \log\lvert y \rvert + C

Correct answerYour answer

e^{y/x} = \log\lvert x \rvert + C

Correct answerYour answer

e^{-x/y} = \log\lvert y \rvert + C

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e^{x/y} = \log\lvert x \rvert + C

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AnswerA · $e^{x/y} = \log\lvert y \rvert + C$

$\int e^{v}\,dv = e^{v}$ and $\int \frac{dy}{y} = \log\lvert y \rvert$ .
Put back $v = \frac{x}{y}$ : $e^{x/y} = \log\lvert y \rvert + C$ .

Worth noting

$e^{y/x} = \log\lvert x \rvert + C$ puts back $v$ as $\frac{y}{x}$ ; with $x = vy$ , $v = \frac{x}{y}$ .

Question 43

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For $x^{2}\frac{dy}{dx} = y^{2} + 2xy$ , which of the following statements are true?

I. $y = vx$ gives $x\frac{dv}{dx} = v^{2} + v$ .

II. The variables separate as $\frac{dv}{v^{2} + 2v} = \frac{dx}{x}$ .

III. The general solution is $y = Cx(x + y)$ .

IV. The line $y = x$ is a solution.

I and III onlyCorrect answerYour answerI, II and III onlyCorrect answerYour answerI and II onlyCorrect answerYour answerIII and IV onlyCorrect answerYour answer

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AnswerA · I and III only

I: $v + x\frac{dv}{dx} = v^{2} + 2v$ , so $x\frac{dv}{dx} = v^{2} + v$ ; II forgets to take away $v$ .
III: $\frac{dv}{v(v + 1)} = \frac{dx}{x}$ gives $\frac{v}{v + 1} = Cx$ , that is $\frac{y}{x + y} = Cx$ . IV: on $y = x$ the left side is $x^{2}$ and the right side $3x^{2}$ , so it fails.

Worth noting

II divides by $F(v) = v^{2} + 2v$ instead of by $F(v) - v$ .

Question 44

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Which of these lines through the origin is a solution of $\frac{dy}{dx} = \frac{2x^{2} + y^{2}}{3xy}$ ?

y = x

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y = 2x

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y = 3x

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y = \frac{x}{2}

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AnswerA · $y = x$

On $y = mx$ , $\frac{dy}{dx} = m$ and the right side is $\frac{2 + m^{2}}{3m}$ ; the line is a solution when these are equal.
$2 + m^{2} = 3m^{2}$ gives $m^{2} = 1$ , so $y = x$ (and $y = -x$ ) are solutions.

Worth noting

$y = 2x$ solves $2 + m^{2} = 3m$ , which forgets to multiply the $m$ on the right by $3m$ .

Question 45

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Consider the differential equation $y\,e^{x/y}\,dx = (x\,e^{x/y} + y)\,dy$ , and put $x = vy$ .

Which separated form does $x = vy$ give?

e^{-v}\,dv = \frac{dy}{y}

Correct answerYour answer

\frac{dv}{v + e^{-v}} = \frac{dy}{y}

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e^{v}\,dv = \frac{dx}{x}

Correct answerYour answer

e^{v}\,dv = \frac{dy}{y}

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AnswerD · $e^{v}\,dv = \frac{dy}{y}$

$\frac{dx}{dy} = v + y\frac{dv}{dy} = v + e^{-v}$ , so $y\frac{dv}{dy} = e^{-v}$ .
Separate: $\frac{dv}{e^{-v}} = e^{v}\,dv = \frac{dy}{y}$ .

Worth noting

$e^{-v}\,dv$ multiplies by $e^{-v}$ where the step divides by it.

Question 46

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$y$ satisfies $\frac{dy}{dx} = \frac{y}{x} + \cos^{2}\frac{y}{x}$ for $x > 0$ , and $y = 0$ when $x = 1$ . What is $y$ when $x = e$ ?

\frac{\pi}{4}

Correct answerYour answer

\frac{e \pi}{4}

Correct answerYour answer

e

Correct answerYour answer

e \tan 1

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AnswerB · $\frac{e \pi}{4}$

$y = vx$ : $x\frac{dv}{dx} = \cos^{2}v$ , so $\sec^{2}v\,dv = \frac{dx}{x}$ and $\tan\frac{y}{x} = \log x + C$ ; $y(1) = 0$ gives $C = 0$ .
At $x = e$ : $\tan\frac{y}{e} = 1$ , so $\frac{y}{e} = \frac{\pi}{4}$ and $y = \frac{e\pi}{4}$ .

Worth noting

$\frac{\pi}{4}$ is $v = \frac{y}{x}$ ; $y$ is $v$ times $x = e$ .

Question 47

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What is the general solution of $y^{2}\frac{dx}{dy} = x^{2} + xy$ ?

\frac{y}{x} = \log\lvert y \rvert + C

Correct answerYour answer

\frac{x}{y} + \log\lvert y \rvert = C

Correct answerYour answer

\frac{y}{x} + \log\lvert x \rvert = C

Correct answerYour answer

\frac{y}{x} + \log\lvert y \rvert = C

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AnswerD · $\frac{y}{x} + \log\lvert y \rvert = C$

$x = vy$ : $v + y\frac{dv}{dy} = v^{2} + v$ , so $\frac{dv}{v^{2}} = \frac{dy}{y}$ .
$-\frac{1}{v} = \log\lvert y \rvert + c$ , and $\frac{1}{v} = \frac{y}{x}$ : $\frac{y}{x} + \log\lvert y \rvert = C$ .

Worth noting

$\frac{y}{x} = \log\lvert y \rvert + C$ loses the minus sign of $\int \frac{dv}{v^{2}} = -\frac{1}{v}$ .

Question 48

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What is the general solution of $\frac{dy}{dx} = \frac{y}{x} + \sin^{2}\frac{y}{x}$ ?

\cot\frac{y}{x} = \log\lvert x \rvert + C

Correct answerYour answer

\tan\frac{y}{x} + \log\lvert x \rvert = C

Correct answerYour answer

\cot\frac{y}{x} + \log\lvert x \rvert = C

Correct answerYour answer

\cot\frac{x}{y} + \log\lvert x \rvert = C

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AnswerC · $\cot\frac{y}{x} + \log\lvert x \rvert = C$

$y = vx$ : $v + x\frac{dv}{dx} = v + \sin^{2}v$ , so $\operatorname{cosec}^{2}v\,dv = \frac{dx}{x}$ .
$-\cot v = \log\lvert x \rvert + C$ , so $\cot\frac{y}{x} + \log\lvert x \rvert = C$ .

Worth noting

$\cot\frac{y}{x} = \log\lvert x \rvert + C$ drops the minus sign: $\int \operatorname{cosec}^{2}v\,dv = -\cot v$ .

Question 49

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Read the two statements and choose the correct option.

I. For $\frac{dx}{dy} = \frac{x}{y} + \tan\frac{x}{y}$ , the substitution $x = vy$ gives $\cot v\,dv = \frac{dy}{y}$ .

II. The general solution of that equation is $\sin\frac{x}{y} = Cy$ .

Only ICorrect answerYour answerOnly IICorrect answerYour answerBoth I and IICorrect answerYour answerNeither I nor IICorrect answerYour answer

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AnswerC · Both I and II

I: $v + y\frac{dv}{dy} = v + \tan v$ , so $\frac{dv}{\tan v} = \cot v\,dv = \frac{dy}{y}$ . True.
II: $\log\lvert \sin v \rvert = \log\lvert y \rvert + c$ , so $\sin\frac{x}{y} = Cy$ . True.

Worth noting

With $x = vy$ the constant multiplies $y$ , the independent variable, not $x$ .

Question 50

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$y$ satisfies $\frac{dy}{dx} = \frac{x + y}{x}$ for $x > 0$ , and $y = 2$ when $x = 1$ . What is $y$ when $x = e$ ?

2 + e

Correct answerYour answer

2 e

Correct answerYour answer

e + 3

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3 e

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AnswerD · $3 e$

The general solution is $y = x\log x + Cx$ ; $y(1) = 2$ gives $C = 2$ .
At $x = e$ : $e \cdot 1 + 2e = 3e$ .

Worth noting

$e + 2$ adds the constant 2 without multiplying it by $x = e$ .

Question 51

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What is the general solution of $\frac{dy}{dx} = \frac{y}{x} + \frac{x}{y}$ ?

y^{2} = x^{2}\log\lvert x \rvert + Cx^{2}

Correct answerYour answer

y^{2} = 2\log\lvert x \rvert + C

Correct answerYour answer

y = 2x\log\lvert x \rvert + Cx

Correct answerYour answer

y^{2} = 2x^{2}\log\lvert x \rvert + Cx^{2}

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AnswerD · $y^{2} = 2x^{2}\log\lvert x \rvert + Cx^{2}$

$y = vx$ : $v + x\frac{dv}{dx} = v + \frac{1}{v}$ , so $v\,dv = \frac{dx}{x}$ .
$\frac{v^{2}}{2} = \log\lvert x \rvert + c$ , so $y^{2} = 2x^{2}\log\lvert x \rvert + Cx^{2}$ .

Worth noting

$y^{2} = x^{2}\log\lvert x \rvert + Cx^{2}$ drops the 2 from $\frac{v^{2}}{2}$ .

Question 52

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What is the general solution of $2xy\frac{dy}{dx} = x^{2} + y^{2}$ ?

x^{2} - y^{2} = Cx

Correct answerYour answer

x^{2} + y^{2} = Cx

Correct answerYour answer

x^{2} - y^{2} = C

Correct answerYour answer

y^{2} = Cx^{2}

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AnswerA · $x^{2} - y^{2} = Cx$

$y = vx$ : $v + x\frac{dv}{dx} = \frac{1 + v^{2}}{2v}$ , so $\frac{2v\,dv}{1 - v^{2}} = \frac{dx}{x}$ .
$-\log\lvert 1 - v^{2} \rvert = \log\lvert x \rvert + c$ , so $x(1 - v^{2}) = C$ : $x^{2} - y^{2} = Cx$ .

Worth noting

$x^{2} - y^{2} = C$ loses the factor $x$ when $1 - v^{2}$ is multiplied out.

Question 53

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Consider the differential equation $x^{2}\frac{dy}{dx} = x^{2} + xy + y^{2}$ , and put $y = vx$ .

The solution curve passes through $(1, 0)$ . What is $y$ when $x = e^{\pi/4}$ ?

\frac{\pi}{4}

Correct answerYour answer

1

Correct answerYour answer

\frac{\pi e^{\frac{\pi}{4}}}{4}

Correct answerYour answer

e^{\frac{\pi}{4}}

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AnswerD · $e^{\frac{\pi}{4}}$

At $(1, 0)$ : $\tan^{-1}0 = \log 1 + C$ , so $C = 0$ and $\tan^{-1}\frac{y}{x} = \log x$ .
At $x = e^{\pi/4}$ : $\tan^{-1}\frac{y}{x} = \frac{\pi}{4}$ , so $\frac{y}{x} = 1$ and $y = e^{\pi/4}$ .

Worth noting

1 is $v = \frac{y}{x}$ ; $y$ is $v$ times $x$ .

Question 54

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What is the general solution of $\frac{dy}{dx} = \frac{y}{x} + \tan\frac{y}{x}$ ?

\cos\frac{y}{x} = Cx

Correct answerYour answer

\tan\frac{y}{x} = Cx

Correct answerYour answer

\sin\frac{y}{x} = C

Correct answerYour answer

\sin\frac{y}{x} = Cx

Correct answerYour answer

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AnswerD · $\sin\frac{y}{x} = Cx$

$y = vx$ : $v + x\frac{dv}{dx} = v + \tan v$ , so $\cot v\,dv = \frac{dx}{x}$ .
$\log\lvert \sin v \rvert = \log\lvert x \rvert + c$ , so $\sin\frac{y}{x} = Cx$ .

Worth noting

$\cos\frac{y}{x} = Cx$ takes $\int \cot v\,dv$ as $\log\lvert \cos v \rvert$ ; that is $-\int \tan v\,dv$ .

Question 55

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What is the general solution of $(x - y)\frac{dy}{dx} = x + y$ ?

\tan^{-1}\frac{y}{x} = \log\sqrt{x^{2} + y^{2}} + C

Correct answerYour answer

\tan^{-1}\frac{y}{x} = \log\lvert x \rvert + C

Correct answerYour answer

\tan^{-1}\frac{x}{y} = \log\sqrt{x^{2} + y^{2}} + C

Correct answerYour answer

\tan^{-1}\frac{y}{x} = -\log\sqrt{x^{2} + y^{2}} + C

Correct answerYour answer

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AnswerA · $\tan^{-1}\frac{y}{x} = \log\sqrt{x^{2} + y^{2}} + C$

$y = vx$ : $v + x\frac{dv}{dx} = \frac{1 + v}{1 - v}$ , so $x\frac{dv}{dx} = \frac{1 + v^{2}}{1 - v}$ and $\frac{1 - v}{1 + v^{2}}dv = \frac{dx}{x}$ .
$\tan^{-1}v - \frac{1}{2}\log(1 + v^{2}) = \log\lvert x \rvert + C$ ; the logs combine to $\log\sqrt{x^{2} + y^{2}}$ .

Worth noting

$\log\lvert x \rvert$ alone forgets the $-\frac{1}{2}\log(1 + v^{2})$ from the $-v$ in the numerator.

Question 56

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Consider the differential equation $x^{2}\frac{dy}{dx} = x^{2} + xy + y^{2}$ , and put $y = vx$ .

What is the general solution of the equation?

\tan^{-1}\frac{y}{x} = \log\lvert x \rvert + C

Correct answerYour answer

\tan^{-1}\frac{x}{y} = \log\lvert x \rvert + C

Correct answerYour answer

\log\left(1 + \frac{y^{2}}{x^{2}}\right) = \log\lvert x \rvert + C

Correct answerYour answer

\tan^{-1}\frac{y}{x} = \log\lvert y \rvert + C

Correct answerYour answer

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AnswerA · $\tan^{-1}\frac{y}{x} = \log\lvert x \rvert + C$

$\int \frac{dv}{1 + v^{2}} = \tan^{-1}v$ and $\int \frac{dx}{x} = \log\lvert x \rvert$ .
$\tan^{-1}\frac{y}{x} = \log\lvert x \rvert + C$ .

Worth noting

$\log\left(1 + v^{2}\right)$ integrates $\frac{1}{1 + v^{2}}$ as if it were $\frac{2v}{1 + v^{2}}$ .

Question 57

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Consider the differential equation $y\,e^{x/y}\,dx = (x\,e^{x/y} + y)\,dy$ , and put $x = vy$ .

The solution curve passes through $(0, 1)$ . What is $x$ when $y = e$ ?

0

Correct answerYour answer

\log 2

Correct answerYour answer

e \log 2

Correct answerYour answer

2 e

Correct answerYour answer

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AnswerC · $e \log 2$

At $(0, 1)$ : $e^{0} = \log 1 + C$ , so $C = 1$ .
At $y = e$ : $e^{x/e} = 1 + 1 = 2$ , so $\frac{x}{e} = \log 2$ and $x = e\log 2$ .

Worth noting

$\log 2$ is $v = \frac{x}{y}$ ; $x$ is $v$ times $y = e$ .

Question 58

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For $x > 0$ , $y$ satisfies $x\frac{dy}{dx} - y = 2\sqrt{x^{2} + y^{2}}$ , and $y = 0$ when $x = 1$ . What is $y$ when $x = 2$ ?

0

Correct answerYour answer

\frac{3}{2}

Correct answerYour answer

\frac{15}{4}

Correct answerYour answer

8

Correct answerYour answer

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AnswerC · $\frac{15}{4}$

The general solution is $y + \sqrt{x^{2} + y^{2}} = Cx^{3}$ ; at $(1, 0)$ , $0 + 1 = C$ , so $C = 1$ .
At $x = 2$ : $\sqrt{4 + y^{2}} = 8 - y$ , so $4 + y^{2} = 64 - 16y + y^{2}$ and $y = \frac{60}{16} = \frac{15}{4}$ .

Worth noting

$\frac{3}{2}$ uses $y + \sqrt{x^{2} + y^{2}} = Cx^{2}$ , which forgets to multiply $v + \sqrt{1 + v^{2}}$ through by $x$ .

Question 59

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For $x > 0$ , what is the general solution of $x\frac{dy}{dx} - y = 2\sqrt{x^{2} + y^{2}}$ ?

y + \sqrt{x^{2} + y^{2}} = Cx^{2}

Correct answerYour answer

y + \sqrt{x^{2} + y^{2}} = Cx^{3}

Correct answerYour answer

y + \sqrt{x^{2} + y^{2}} = Cx

Correct answerYour answer

y - \sqrt{x^{2} + y^{2}} = Cx^{3}

Correct answerYour answer

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AnswerB · $y + \sqrt{x^{2} + y^{2}} = Cx^{3}$

$y = vx$ : $x\frac{dv}{dx} = 2\sqrt{1 + v^{2}}$ , so $\frac{dv}{\sqrt{1 + v^{2}}} = \frac{2\,dx}{x}$ .
$\log\left\lvert v + \sqrt{1 + v^{2}} \right\rvert = 2\log x + c$ , so $v + \sqrt{1 + v^{2}} = Cx^{2}$ .
Multiply by $x$ : $y + \sqrt{x^{2} + y^{2}} = Cx^{3}$ .

Worth noting

$Cx^{2}$ stops at $v + \sqrt{1 + v^{2}} = Cx^{2}$ ; the left side is still divided by $x$ .

Question 60

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What is the general solution of $(x^{2} - xy)\,dy = (xy + y^{2})\,dx$ ?

\frac{x}{y} + \log\lvert y \rvert = C

Correct answerYour answer

\frac{x}{y} - \log\lvert xy \rvert = C

Correct answerYour answer

\frac{y}{x} + \log\lvert xy \rvert = C

Correct answerYour answer

\frac{x}{y} + \log\lvert xy \rvert = C

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AnswerD · $\frac{x}{y} + \log\lvert xy \rvert = C$

$y = vx$ : $v + x\frac{dv}{dx} = \frac{v(1 + v)}{1 - v}$ , so $x\frac{dv}{dx} = \frac{2v^{2}}{1 - v}$ and $\left(\frac{1}{2v^{2}} - \frac{1}{2v}\right)dv = \frac{dx}{x}$ .
$-\frac{1}{2v} - \frac{1}{2}\log\lvert v \rvert = \log\lvert x \rvert + c$ ; times $-2$ : $\frac{1}{v} + \log\lvert v \rvert + 2\log\lvert x \rvert = C$ .
$\log\left\lvert \frac{y}{x} \right\rvert + 2\log\lvert x \rvert = \log\lvert xy \rvert$ , so $\frac{x}{y} + \log\lvert xy \rvert = C$ .

Worth noting

$\frac{x}{y} + \log\lvert y \rvert = C$ drops the 2 when $\log\lvert x \rvert$ is multiplied by $-2$ .

Linear equations and the integrating factor

29 questions

Question 61

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What is the integrating factor of $\frac{dy}{dx} + \frac{2}{x}y = x$ ?

2 \log x

Correct answerYour answer

x

Correct answerYour answer

x^{2}

Correct answerYour answer

e^{2 x}

Correct answerYour answer

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AnswerC · $x^{2}$

In the form $\frac{dy}{dx} + Py = Q$ , $P = \frac{2}{x}$ .
The factor is $e^{\int P\,dx} = x^{2}$ .

Worth noting

$2\log x$ is $\int P\,dx$ ; the factor is $e$ raised to it, $x^{2}$ .

Question 62

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Which of these differential equations is linear in $y$ ?

\frac{dy}{dx} + y^{2} = x

Correct answerYour answer

y\frac{dy}{dx} + x = 0

Correct answerYour answer

\frac{dy}{dx} = \sin y

Correct answerYour answer

\frac{dy}{dx} + xy = \sin x

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AnswerD · $\frac{dy}{dx} + xy = \sin x$

Linear in $y$ means the form $\frac{dy}{dx} + Py = Q$ with $P$ , $Q$ functions of $x$ only.
Only $\frac{dy}{dx} + xy = \sin x$ fits, with $P = x$ and $Q = \sin x$ .

Worth noting

$y\frac{dy}{dx}$ multiplies $y$ by its derivative, which a linear equation never does.

Question 63

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What is the integrating factor of $\frac{dy}{dx} + y = e^{-x}$ ?

e^{- x}

Correct answerYour answer

x

Correct answerYour answer

e^{x}

Correct answerYour answer

1

Correct answerYour answer

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AnswerC · $e^{x}$

In the form $\frac{dy}{dx} + Py = Q$ , $P = 1$ .
The factor is $e^{\int P\,dx} = e^{x}$ .

Worth noting

$e^{-x}$ is $Q$ , the right-hand side; the factor comes from $P = 1$ .

Question 64

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What is the integrating factor of $\frac{dy}{dx} + 3y = x$ ?

3 x

Correct answerYour answer

e^{x}

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e^{3}

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e^{3 x}

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AnswerD · $e^{3 x}$

In the form $\frac{dy}{dx} + Py = Q$ , $P = 3$ .
The factor is $e^{\int P\,dx} = e^{3 x}$ .

Worth noting

$3x$ is $\int P\,dx$ ; the factor is $e^{3x}$ .

Question 65

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Which of these is the standard form of a differential equation that is linear in $x$ ?

\frac{dx}{dy} + P_{1}x = Q_{1}

, with

P_{1}

Q_{1}

functions of

x

Correct answerYour answer

\frac{dx}{dy} + P_{1}x = Q_{1}

, with

P_{1}

Q_{1}

functions of

y

Correct answerYour answer

\frac{dy}{dx} + P_{1}x = Q_{1}

, with

P_{1}

Q_{1}

functions of

y

Correct answerYour answer

\frac{dx}{dy} + P_{1}x^{2} = Q_{1}

, with

P_{1}

Q_{1}

functions of

y

Correct answerYour answer

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AnswerB · $\frac{dx}{dy} + P_{1}x = Q_{1}$ , with $P_{1}$ , $Q_{1}$ functions of $y$

With $x$ as the unknown, $y$ is the independent variable, so the derivative is $\frac{dx}{dy}$ .
$x$ appears only to the first power, and $P_{1}$ , $Q_{1}$ may depend on $y$ alone.

Worth noting

$P_{1}$ and $Q_{1}$ cannot contain $x$ , the unknown; they are functions of $y$ .

Question 66

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What is the integrating factor of $\frac{dx}{dy} + 2x = y^{2}$ ?

e^{2 x}

Correct answerYour answer

2 y

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e^{- 2 y}

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e^{2 y}

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AnswerD · $e^{2 y}$

The equation is linear in $x$ with $P_{1} = 2$ , a constant function of $y$ .
The factor is $e^{\int 2\,dy} = e^{2y}$ .

Worth noting

$e^{2x}$ integrates with respect to $x$ ; here the variable of integration is $y$ .

Question 67

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What is the integrating factor of $\frac{dy}{dx} = x + y$ ?

e^{- x}

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e^{x}

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e^{- y}

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x

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AnswerA · $e^{- x}$

In standard form, $\frac{dy}{dx} - y = x$ , so $P = -1$ .
The factor is $e^{\int -1\,dx} = e^{-x}$ .

Worth noting

$e^{x}$ reads $P = +1$ off $\frac{dy}{dx} = x + y$ ; moving $y$ to the left makes it $-1$ .

Question 68

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Assertion (A): The integrating factor of $\frac{dy}{dx} + Py = Q$ is $e^{\int P\,dx}$ .

Reason (R): Multiplying by $e^{\int P\,dx}$ turns the left side into $\frac{d}{dx}\left(y\,e^{\int P\,dx}\right)$ .

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AnswerA · Both A and R are true, and R is the correct explanation of A.

A is true, and R is the reason: the product rule gives $e^{\int P\,dx}(y' + Py)$ .
Then integrating both sides solves the equation.

Question 69

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What is the integrating factor of $\frac{dx}{dy} + \frac{x}{y} = y^{2}$ , linear in $x$ ?

x

Correct answerYour answer

\log y

Correct answerYour answer

y

Correct answerYour answer

y^{2}

Correct answerYour answer

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AnswerC · $y$

Here $x$ is the unknown: $\frac{dx}{dy} + Px = Q$ with $P = \frac{1}{y}$ .
The factor is $e^{\int \frac{1}{y}dy} = y$ .

Worth noting

$x$ is the unknown here; the factor is a function of $y$ .

Question 70

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What is the integrating factor of $\frac{dy}{dx} - y\tan x = \sin x$ ?

\sec x

Correct answerYour answer

\cos x

Correct answerYour answer

- \cos x

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e^{\tan x}

Correct answerYour answer

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AnswerB · $\cos x$

In the form $\frac{dy}{dx} + Py = Q$ , $P = - \tan x$ .
The factor is $e^{\int P\,dx} = \cos x$ .

Worth noting

$\sec x$ is the factor for $+y\tan x$ ; with the minus sign, $\int -\tan x\,dx = \log \cos x$ .

Question 71

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What is the integrating factor of $x\frac{dy}{dx} - y = x^{2}$ ?

x

Correct answerYour answer

\frac{1}{x}

Correct answerYour answer

- \log x

Correct answerYour answer

e^{- x}

Correct answerYour answer

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AnswerB · $\frac{1}{x}$

In the form $\frac{dy}{dx} + Py = Q$ , $P = - \frac{1}{x}$ .
The factor is $e^{\int P\,dx} = \frac{1}{x}$ .

Worth noting

$x$ skips dividing through by $x$ first: then $P = -\frac{1}{x}$ and the factor is $\frac{1}{x}$ .

Question 72

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Read the two statements and choose the correct option.

I. $\frac{dy}{dx} + \frac{y}{x} = \sin x$ has the integrating factor $x$ .

II. If $I$ is an integrating factor, so is $5I$ .

Only ICorrect answerYour answerOnly IICorrect answerYour answerBoth I and IICorrect answerYour answerNeither I nor IICorrect answerYour answer

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AnswerC · Both I and II

I: $e^{\int \frac{dx}{x}} = e^{\log x} = x$ . True.
II: multiplying by $5I$ only scales both sides by 5, and the left side is still an exact derivative. True.

Question 73

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What is the integrating factor of $y\frac{dx}{dy} - x = 2y^{3}$ ?

y

Correct answerYour answer

\frac{1}{x}

Correct answerYour answer

e^{-y}

Correct answerYour answer

\frac{1}{y}

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AnswerD · $\frac{1}{y}$

Divide by $y$ : $\frac{dx}{dy} - \frac{1}{y}x = 2y^{2}$ , linear in $x$ with $P = -\frac{1}{y}$ .
$e^{-\int \frac{dy}{y}} = e^{-\log y} = \frac{1}{y}$ .

Worth noting

$y$ is the factor for $+\frac{x}{y}$ ; the minus sign turns it into $\frac{1}{y}$ .

Question 74

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What is the integrating factor of $\frac{dy}{dx} + y\cot x = 2\cos x$ ?

\sin x

Correct answerYour answer

\cos x

Correct answerYour answer

\log{\left(\sin x \right)}

Correct answerYour answer

\frac{1}{\sin x}

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AnswerA · $\sin x$

In the form $\frac{dy}{dx} + Py = Q$ , $P = \cot x$ .
The factor is $e^{\int P\,dx} = \sin x$ .

Worth noting

$\log \sin x$ is $\int \cot x\,dx$ ; the factor is $e$ raised to it.

Question 75

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What is the integrating factor of $(1 + x^{2})\frac{dy}{dx} + 2xy = 1$ ?

\log{\left(x^{2} + 1 \right)}

Correct answerYour answer

2 x

Correct answerYour answer

x^{2} + 1

Correct answerYour answer

e^{x^{2}}

Correct answerYour answer

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AnswerC · $x^{2} + 1$

In the form $\frac{dy}{dx} + Py = Q$ , $P = \frac{2 x}{x^{2} + 1}$ .
The factor is $e^{\int P\,dx} = x^{2} + 1$ .

Worth noting

$e^{x^{2}}$ uses $P = 2x$ before dividing through by $1 + x^{2}$ .

Question 76

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In the standard form $\frac{dy}{dx} + Py = Q$ , what are $P$ and $Q$ for $x\frac{dy}{dx} - 3y = x^{4}e^{x}$ ?

P = -3

Q = x^{4}e^{x}

Correct answerYour answer

P = \frac{3}{x}

Q = x^{3}e^{x}

Correct answerYour answer

P = -\frac{3}{x}

Q = x^{3}e^{x}

Correct answerYour answer

P = -\frac{3}{x}

Q = x^{4}e^{x}

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AnswerC · $P = -\frac{3}{x}$ , $Q = x^{3}e^{x}$

Divide through by $x$ , the coefficient of $\frac{dy}{dx}$ : $\frac{dy}{dx} - \frac{3}{x}y = x^{3}e^{x}$ .
So $P = -\frac{3}{x}$ and $Q = x^{3}e^{x}$ .

Worth noting

$P = -3$ , $Q = x^{4}e^{x}$ reads the equation before dividing by $x$ , the coefficient of $\frac{dy}{dx}$ .

Question 77

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For $x > 0$ , the integrating factor of $\frac{dy}{dx} + Py = Q$ is $x^{3}$ . What is $P$ ?

3 x^{2}

Correct answerYour answer

3 \log x

Correct answerYour answer

\frac{1}{x^{3}}

Correct answerYour answer

\frac{3}{x}

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AnswerD · $\frac{3}{x}$

$e^{\int P\,dx} = x^{3}$ means $\int P\,dx = \log x^{3} = 3\log x$ .
Differentiate: $P = \frac{3}{x}$ .

Worth noting

$3x^{2}$ differentiates the factor itself; $\int P\,dx$ is the logarithm of the factor.

Question 78

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For $x > 0$ , $\frac{dy}{dx} - \frac{2}{x}y = x$ is multiplied by its integrating factor. Which exact derivative does the left side become?

\frac{d}{dx}\left(x^{2}y\right)

Correct answerYour answer

\frac{d}{dx}\left(\frac{y}{x}\right)

Correct answerYour answer

\frac{d}{dx}\left(\frac{y}{x^{2}}\right)

Correct answerYour answer

\frac{d}{dx}\left(e^{-2x}y\right)

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AnswerC · $\frac{d}{dx}\left(\frac{y}{x^{2}}\right)$

$P = -\frac{2}{x}$ , so the factor is $e^{-2\log x} = \frac{1}{x^{2}}$ .
By the quotient rule, $\frac{1}{x^{2}}\frac{dy}{dx} - \frac{2}{x^{3}}y = \frac{d}{dx}\left(\frac{y}{x^{2}}\right)$ .

Worth noting

$x^{2}y$ uses the factor for $+\frac{2}{x}$ ; the minus sign turns it into $\frac{1}{x^{2}}$ .

Question 79

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Which of the following statements are true?

I. An integrating factor $e^{\int P\,dx}$ is never 0.

II. If $P$ is a constant $a$ , the integrating factor is $e^{ax}$ .

III. The integrating factor depends on $Q$ as well as on $P$ .

IV. For $x > 0$ , $\frac{dy}{dx} + \frac{y}{x} = Q$ and $x\frac{dy}{dx} + y = xQ$ have the same integrating factor.

All four: I, II, III and IVCorrect answerYour answerI, II and IV onlyCorrect answerYour answerI and II onlyCorrect answerYour answerII and IV onlyCorrect answerYour answer

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AnswerB · I, II and IV only

I: a power of $e$ is always positive. II: $e^{\int a\,dx} = e^{ax}$ . IV: dividing the second equation by $x$ gives the first, so both have $P = \frac{1}{x}$ and factor $x$ .
III is false: the factor is $e^{\int P\,dx}$ , built from $P$ alone.

Worth noting

$Q$ enters only after the factor is found, in $\int Q \cdot e^{\int P\,dx}\,dx$ .

Question 80

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Which of these equations is linear when $x$ is taken as the unknown, a function of $y$ ?

\frac{dy}{dx} + y = x^{2}

Correct answerYour answer

\frac{dy}{dx} = \frac{1}{x + y^{2}}

Correct answerYour answer

\frac{dy}{dx} = \frac{1}{x^{2} + y}

Correct answerYour answer

\frac{dy}{dx} = \frac{y}{x^{2} + y}

Correct answerYour answer

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AnswerB · $\frac{dy}{dx} = \frac{1}{x + y^{2}}$

Flip each one to $\frac{dx}{dy}$ : $\frac{dy}{dx} = \frac{1}{x + y^{2}}$ gives $\frac{dx}{dy} - x = y^{2}$ , linear in $x$ with $P_{1} = -1$ .
The others give $\frac{dx}{dy} = \frac{1}{x^{2} - y}$ , $x^{2} + y$ and $\frac{x^{2} + y}{y}$ , each with $x^{2}$ in it.

Worth noting

$\frac{dy}{dx} + y = x^{2}$ is linear in $y$ ; flipped, $\frac{dx}{dy} = \frac{1}{x^{2} - y}$ is not linear in $x$ .

Question 81

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Read the two statements and choose the correct option.

I. $\frac{dy}{dx} + \frac{y}{x} = y^{2}$ is a linear differential equation in $y$ .

II. $\frac{dy}{dx} = \frac{y}{x} + 2$ is both homogeneous and linear.

Only ICorrect answerYour answerOnly IICorrect answerYour answerBoth I and IICorrect answerYour answerNeither I nor IICorrect answerYour answer

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AnswerB · Only II

I: the $y^{2}$ on the right makes it non-linear in $y$ . False.
II: the right side is $v + 2$ , a function of $\frac{y}{x}$ , and $\frac{dy}{dx} - \frac{1}{x}y = 2$ is linear. True.

Worth noting

One equation can fit both methods: being homogeneous and being linear are separate tests.

Question 82

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What is the integrating factor of $\cos x\frac{dy}{dx} + y = 1$ , for $0 < x < \frac{\pi}{2}$ ?

\sec x + \tan x

Correct answerYour answer

\sec x

Correct answerYour answer

e^{\sec x}

Correct answerYour answer

e^{\tan x}

Correct answerYour answer

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AnswerA · $\sec x + \tan x$

Divide by $\cos x$ : $\frac{dy}{dx} + y\sec x = \sec x$ , so $P = \sec x$ .
$\int \sec x\,dx = \log(\sec x + \tan x)$ , so the factor is $\sec x + \tan x$ .

Worth noting

$\sec x$ is $P$ itself, not $e^{\int P\,dx}$ .

Question 83

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Assertion (A): For $x > -1$ , the integrating factor of $\frac{dy}{dx} - \frac{y}{x + 1} = 1$ is $x + 1$ .

Reason (R): For $x > -1$ , $e^{-\log(x + 1)} = \frac{1}{x + 1}$ .

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AnswerD · A is false, but R is true.

$P = -\frac{1}{x + 1}$ , so $\int P\,dx = -\log(x + 1)$ and the factor is $e^{-\log(x + 1)} = \frac{1}{x + 1}$ .
So A is false and R is true; R in fact gives the correct factor.

Worth noting

$x + 1$ is the factor for $+\frac{y}{x + 1}$ ; the minus sign inverts it.

Question 84

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For $x > 0$ and $0 < y < \pi$ , match each equation in List I with its integrating factor in List II.

List I	List II
A. $x\frac{dy}{dx} + 4y = 1$	I. $e^{- 3 x}$
B. $\frac{dy}{dx} - 3y = x$	II. $\sin y$
C. $(x + 1)\frac{dy}{dx} + y = 1$	III. $x^{4}$
D. $\frac{dx}{dy} + x\cot y = 1$	IV. $x + 1$

A-III, B-I, C-IV, D-IICorrect answerYour answerA-I, B-III, C-IV, D-IICorrect answerYour answerA-III, B-I, C-II, D-IVCorrect answerYour answerA-I, B-III, C-II, D-IVCorrect answerYour answer

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AnswerA · A-III, B-I, C-IV, D-II

A: divide by $x$ , $P = \frac{4}{x}$ , factor $x^{4}$ . B: $P = -3$ , factor $e^{-3x}$ .
C: $P = \frac{1}{x + 1}$ , factor $x + 1$ . D: linear in $x$ with $P_{1} = \cot y$ , factor $e^{\log \sin y} = \sin y$ .

Question 85

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What is the integrating factor of $x\log x\frac{dy}{dx} + y = 2\log x$ ?

x \log x

Correct answerYour answer

\log x

Correct answerYour answer

x

Correct answerYour answer

\frac{1}{\log x}

Correct answerYour answer

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AnswerB · $\log x$

In the form $\frac{dy}{dx} + Py = Q$ , $P = \frac{1}{x \log x}$ .
The factor is $e^{\int P\,dx} = \log x$ .

Worth noting

$x\log x$ is the coefficient of $\frac{dy}{dx}$ ; dividing by it gives $P = \frac{1}{x\log x}$ .

Question 86

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For $x > 0$ , which of these equations has the integrating factor $x^{2}$ ?

x\frac{dy}{dx} - 2y = e^{x}

Correct answerYour answer

x\frac{dy}{dx} + 2y = e^{x}

Correct answerYour answer

\frac{dy}{dx} + 2xy = e^{x}

Correct answerYour answer

x^{2}\frac{dy}{dx} + 2y = e^{x}

Correct answerYour answer

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AnswerB · $x\frac{dy}{dx} + 2y = e^{x}$

$x\frac{dy}{dx} + 2y = e^{x}$ , divided by $x$ , has $P = \frac{2}{x}$ , so its factor is $e^{2\log x} = x^{2}$ .
The others have $P = -\frac{2}{x}$ , $2x$ and $\frac{2}{x^{2}}$ , with factors $\frac{1}{x^{2}}$ , $e^{x^{2}}$ and $e^{-2/x}$ .

Worth noting

$x\frac{dy}{dx} - 2y = e^{x}$ has $P = -\frac{2}{x}$ , whose factor is $\frac{1}{x^{2}}$ , not $x^{2}$ .

Question 87

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Treating $x$ as the unknown, what is the integrating factor of $(1 + y^{2})\,dx = (2\tan^{-1}y - x)\,dy$ ?

e^{-\tan^{-1}y}

Correct answerYour answer

e^{\tan^{-1}x}

Correct answerYour answer

e^{\tan^{-1}y}

Correct answerYour answer

1 + y^{2}

Correct answerYour answer

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AnswerC · $e^{\tan^{-1}y}$

$\frac{dx}{dy} = \frac{2\tan^{-1}y - x}{1 + y^{2}}$ , so $\frac{dx}{dy} + \frac{x}{1 + y^{2}} = \frac{2\tan^{-1}y}{1 + y^{2}}$ .
$P_{1} = \frac{1}{1 + y^{2}}$ , so the factor is $e^{\int \frac{dy}{1 + y^{2}}} = e^{\tan^{-1}y}$ .

Worth noting

$e^{-\tan^{-1}y}$ keeps $-x$ on the right; moved to the left it becomes $+\frac{x}{1 + y^{2}}$ .

Question 88

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What is the integrating factor of $x\frac{dy}{dx} + (1 + x)y = x^{2}$ , for $x > 0$ ?

x + e^{x}

Correct answerYour answer

e^{x}

Correct answerYour answer

e^{x + 1}

Correct answerYour answer

x e^{x}

Correct answerYour answer

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AnswerD · $x e^{x}$

Divide by $x$ : $P = \frac{1 + x}{x} = \frac{1}{x} + 1$ , so $\int P\,dx = \log x + x$ .
The factor is $e^{\log x + x} = e^{\log x} \cdot e^{x} = x\,e^{x}$ .

Worth noting

$x + e^{x}$ splits $e^{\log x + x}$ into a sum; $e$ raised to a sum is a product.

Question 89

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What is the integrating factor of $(1 - x^{2})\frac{dy}{dx} - xy = 1$ , for $-1 < x < 1$ ?

\sqrt{1 - x^{2}}

Correct answerYour answer

\frac{1}{\sqrt{1 - x^{2}}}

Correct answerYour answer

1 - x^{2}

Correct answerYour answer

e^{- \frac{x^{2}}{2}}

Correct answerYour answer

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AnswerA · $\sqrt{1 - x^{2}}$

Divide by $1 - x^{2}$ : $P = -\frac{x}{1 - x^{2}}$ .
$\int -\frac{x}{1 - x^{2}}\,dx = \frac{1}{2}\log(1 - x^{2})$ , so the factor is $\sqrt{1 - x^{2}}$ .

Worth noting

$\frac{1}{\sqrt{1 - x^{2}}}$ gets the sign of the logarithm wrong: the derivative of $1 - x^{2}$ is $-2x$ , which absorbs the minus in $P$ .

Solving linear equations

31 questions

Question 90

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What is the general solution of $\frac{dy}{dx} + 2y = 4$ ?

y = 4 + Ce^{-2x}

Correct answerYour answer

y = 2 + Ce^{2x}

Correct answerYour answer

y = 2x + C

Correct answerYour answer

y = 2 + Ce^{-2x}

Correct answerYour answer

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AnswerD · $y = 2 + Ce^{-2x}$

The factor is $e^{2x}$ : $\frac{d}{dx}(ye^{2x}) = 4e^{2x}$ , so $ye^{2x} = 2e^{2x} + C$ .
$y = 2 + Ce^{-2x}$ .

Worth noting

$4 + Ce^{-2x}$ forgets to divide 4 by $P = 2$ ; the steady value is where $2y = 4$ .

Question 91

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Consider the linear differential equation $\frac{dy}{dx} + 2xy = 2x$ .

What is its integrating factor?

e^{2 x}

Correct answerYour answer

x^{2}

Correct answerYour answer

e^{x^{2}}

Correct answerYour answer

e^{- x^{2}}

Correct answerYour answer

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AnswerC · $e^{x^{2}}$

$P = 2x$ , so $\int P\,dx = x^{2}$ .
The factor is $e^{x^{2}}$ .

Worth noting

$e^{2x}$ puts $P$ itself in the exponent instead of $\int P\,dx$ .

Question 92

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Consider the linear differential equation $\frac{dy}{dx} + 2xy = 2x$ .

If $y = 3$ when $x = 0$ , what is $C$ ?

-2

Correct answerYour answer

2

Correct answerYour answer

3

Correct answerYour answer

4

Correct answerYour answer

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AnswerB · $2$

At $x = 0$ : $3 = 1 + C \cdot e^{0}$ .
$C = 2$ .

Worth noting

3 takes $C$ as the starting value of $y$ , forgetting the 1.

Question 93

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What is the general solution of $\frac{dy}{dx} - y = 2$ ?

y = Ce^{x} - 2

Correct answerYour answer

y = Ce^{x} + 2

Correct answerYour answer

y = Ce^{-x} - 2

Correct answerYour answer

y = 2x + Ce^{x}

Correct answerYour answer

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AnswerA · $y = Ce^{x} - 2$

The factor is $e^{-x}$ : $\frac{d}{dx}(ye^{-x}) = 2e^{-x}$ , so $ye^{-x} = -2e^{-x} + C$ .
$y = Ce^{x} - 2$ .

Worth noting

$Ce^{x} + 2$ loses the minus sign from $\int 2e^{-x}\,dx = -2e^{-x}$ .

Question 94

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Every solution of $\frac{dy}{dx} + 4y = 8$ approaches the same value as $x \to \infty$ . What is that value?

\frac{1}{2}

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2

Correct answerYour answer

4

Correct answerYour answer

8

Correct answerYour answer

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AnswerB · $2$

The factor $e^{4x}$ gives $ye^{4x} = 2e^{4x} + C$ , so $y = 2 + Ce^{-4x}$ .
As $x \to \infty$ , $Ce^{-4x} \to 0$ , so $y \to \frac{8}{4} = 2$ .

Worth noting

8 is $Q$ itself; the steady value is where $4y = 8$ .

Question 95

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Multiplying a linear equation by its integrating factor $x^{2}$ gives $\frac{d}{dx}\left(x^{2}y\right) = 3x^{2}$ . What is the general solution?

y = x + C

Correct answerYour answer

y = x^{3} + C

Correct answerYour answer

y = \frac{x}{3} + \frac{C}{x^{2}}

Correct answerYour answer

y = x + \frac{C}{x^{2}}

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AnswerD · $y = x + \frac{C}{x^{2}}$

Integrate: $x^{2}y = \int 3x^{2}\,dx = x^{3} + C$ .
Divide by $x^{2}$ : $y = x + \frac{C}{x^{2}}$ .

Worth noting

$x + C$ divides $x^{3}$ by $x^{2}$ but leaves the constant undivided.

Question 96

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A savings account earns interest at 5% a year, compounded continuously, and money is paid in continuously at the rate of ₹12,000 a year. The balance $A$ (in rupees) after $t$ years satisfies $\frac{dA}{dt} = \frac{A}{20} + 12{,}000$ .

What is the integrating factor of this equation?

e^{t/20}

Correct answerYour answer

e^{-20t}

Correct answerYour answer

e^{-t/20}

Correct answerYour answer

e^{-t}

Correct answerYour answer

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AnswerC · $e^{-t/20}$

In standard form, $\frac{dA}{dt} - \frac{1}{20}A = 12{,}000$ , so $P = -\frac{1}{20}$ .
The factor is $e^{\int -\frac{1}{20}\,dt} = e^{-t/20}$ .

Worth noting

$e^{t/20}$ reads $P = +\frac{1}{20}$ ; moving $\frac{A}{20}$ to the left makes it negative.

Question 97

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What is the general solution for $A$ ?

A = Ce^{t/20} + 2{,}40{,}000

Correct answerYour answer

A = Ce^{-t/20} - 2{,}40{,}000

Correct answerYour answer

A = Ce^{t/20} - 600

Correct answerYour answer

A = Ce^{t/20} - 2{,}40{,}000

Correct answerYour answer

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AnswerD · $A = Ce^{t/20} - 2{,}40{,}000$

$\frac{d}{dt}\left(Ae^{-t/20}\right) = 12{,}000e^{-t/20}$ , so $Ae^{-t/20} = -2{,}40{,}000e^{-t/20} + C$ .
Multiply by $e^{t/20}$ : $A = Ce^{t/20} - 2{,}40{,}000$ .

Worth noting

$+2{,}40{,}000$ loses the minus sign of $\int e^{-t/20}\,dt = -20e^{-t/20}$ .

Question 98

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$y$ satisfies $\frac{dy}{dx} + y = 1$ and $y = 0$ when $x = 0$ . What is $y$ when $x = \log 2$ ?

\frac{1}{2}

Correct answerYour answer

1

Correct answerYour answer

\frac{3}{2}

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2

Correct answerYour answer

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AnswerA · $\frac{1}{2}$

The factor $e^{x}$ gives $ye^{x} = e^{x} + C$ , so $y = 1 + Ce^{-x}$ ; $y(0) = 0$ gives $C = -1$ .
$y = 1 - e^{-\log 2} = 1 - \frac{1}{2} = \frac{1}{2}$ .

Worth noting

$\frac{3}{2}$ takes $C = +1$ ; the condition $y(0) = 0$ needs $1 + C = 0$ .

Question 99

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Which of the following statements are true?

I. A linear equation $\frac{dy}{dx} + Py = Q$ has general solution $y\,e^{\int P\,dx} = \int Q\,e^{\int P\,dx}\,dx + C$ .

II. The constant $C$ is added after integrating the right side, not to the factor.

III. The integrating factor needs its own constant of integration.

IV. If $Q = 0$ , the equation also has separable variables.

All four: I, II, III and IVCorrect answerYour answerI, II and IV onlyCorrect answerYour answerI and II onlyCorrect answerYour answerI and IV onlyCorrect answerYour answer

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AnswerB · I, II and IV only

I is the solution formula, II is how it is used, and IV holds: $\frac{dy}{y} = -P\,dx$ .
III is false: any one integrating factor works, so its constant is left out.

Worth noting

A constant in the factor would only scale both sides of the equation.

Question 100

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What is the general solution of $\frac{dy}{dx} + y = e^{-x}$ ?

y = Ce^{-x}

Correct answerYour answer

y = (x + C)e^{-x}

Correct answerYour answer

y = (x + C)e^{x}

Correct answerYour answer

y = x + Ce^{-x}

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AnswerB · $y = (x + C)e^{-x}$

The factor is $e^{x}$ : $\frac{d}{dx}(ye^{x}) = e^{x} \cdot e^{-x} = 1$ .
$ye^{x} = x + C$ , so $y = (x + C)e^{-x}$ .

Worth noting

$Ce^{-x}$ solves $\frac{dy}{dx} + y = 0$ , without the right-hand side.

Question 101

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Consider the linear differential equation $\frac{dy}{dx} + 2xy = 2x$ .

With $y = 3$ when $x = 0$ , what is $y$ when $x = 1$ ?

\frac{2}{e}

Correct answerYour answer

\frac{3}{e}

Correct answerYour answer

\frac{2}{e} + 1

Correct answerYour answer

1 + 2 e

Correct answerYour answer

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AnswerC · $\frac{2}{e} + 1$

$y = 1 + 2e^{-x^{2}}$ .
At $x = 1$ : $1 + \frac{2}{e}$ .

Worth noting

$1 + 2e$ drops the minus sign in the exponent $-x^{2}$ .

Question 102

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What is the general solution of $\frac{dy}{dx} + \frac{y}{x} = x^{2}$ ?

y = \frac{x^{3}}{3} + \frac{C}{x}

Correct answerYour answer

y = \frac{x^{3}}{4} + \frac{C}{x}

Correct answerYour answer

y = \frac{x^{3}}{4} + Cx

Correct answerYour answer

y = \frac{x^{4}}{4} + C

Correct answerYour answer

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AnswerB · $y = \frac{x^{3}}{4} + \frac{C}{x}$

The factor is $x$ : $\frac{d}{dx}(xy) = x^{3}$ .
$xy = \frac{x^{4}}{4} + C$ , so $y = \frac{x^{3}}{4} + \frac{C}{x}$ .

Worth noting

$\frac{x^{3}}{3}$ integrates $Q = x^{2}$ without first multiplying it by the factor $x$ .

Question 103

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What is the general solution of $x\frac{dy}{dx} - 2y = x^{3}$ ?

y = x^{3} + Cx^{2}

Correct answerYour answer

y = x^{3} + C

Correct answerYour answer

y = \frac{x^{3}}{3} + Cx^{2}

Correct answerYour answer

y = x^{3} + \frac{C}{x^{2}}

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AnswerA · $y = x^{3} + Cx^{2}$

Divide by $x$ : $\frac{dy}{dx} - \frac{2}{x}y = x^{2}$ , with factor $\frac{1}{x^{2}}$ .
$\frac{y}{x^{2}} = \int 1\,dx = x + C$ , so $y = x^{3} + Cx^{2}$ .

Worth noting

$x^{3} + C$ forgets to multiply the constant by $x^{2}$ .

Question 104

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Consider the linear differential equation $\frac{dy}{dx} + 2xy = 2x$ .

What is its general solution?

y = 1 + Ce^{x^{2}}

Correct answerYour answer

y = 1 + Ce^{-x^{2}}

Correct answerYour answer

y = x^{2} + Ce^{-x^{2}}

Correct answerYour answer

y = Ce^{-x^{2}}

Correct answerYour answer

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AnswerB · $y = 1 + Ce^{-x^{2}}$

$\frac{d}{dx}\left(ye^{x^{2}}\right) = 2xe^{x^{2}}$ , so $ye^{x^{2}} = e^{x^{2}} + C$ .
$y = 1 + Ce^{-x^{2}}$ .

Worth noting

$Ce^{-x^{2}}$ solves the equation with the right side 0, dropping the 1 that $2x$ adds.

Question 105

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The balance is ₹0 at $t = 0$ . What is the balance after 20 years?

₹

600(e - 1)

Correct answerYour answer₹

2{,}40{,}000

Correct answerYour answer₹

2{,}40{,}000(e - 1)

Correct answerYour answer₹

2{,}40{,}000e

Correct answerYour answer

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AnswerC · ₹ $2{,}40{,}000(e - 1)$

$A(0) = C - 2{,}40{,}000 = 0$ , so $C = 2{,}40{,}000$ and $A = 2{,}40{,}000\left(e^{t/20} - 1\right)$ .
At $t = 20$ : $A = 2{,}40{,}000(e - 1)$ .

Worth noting

₹2,40,000 is only the money paid in, $12{,}000 \times 20$ , with no interest.

Question 106

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For $\frac{dy}{dx} + \frac{y}{x} = 3$ , with $x > 0$ , which of the following statements are true?

I. Its integrating factor is $x$ .

II. Its general solution is $y = 3x + \frac{C}{x}$ .

III. $y = 3x$ is one of its solutions.

IV. Its solution through $(1, 2)$ is $y = \frac{3x}{2} + \frac{1}{2x}$ .

I, II and IV onlyCorrect answerYour answerI and II onlyCorrect answerYour answerI, III and IV onlyCorrect answerYour answerI and IV onlyCorrect answerYour answer

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AnswerD · I and IV only

I: $e^{\int \frac{dx}{x}} = x$ , and $\frac{d}{dx}(xy) = 3x$ gives $xy = \frac{3x^{2}}{2} + C$ , so $y = \frac{3x}{2} + \frac{C}{x}$ ; II has lost the $\frac{1}{2}$ .
III: on $y = 3x$ the left side is $3 + 3 = 6$ , not 3. IV: $\frac{3}{2} + C = 2$ gives $C = \frac{1}{2}$ .

Worth noting

II integrates $3x$ as $3x^{2}$ ; $\int 3x\,dx = \frac{3x^{2}}{2}$ .

Question 107

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What is the general solution of $\frac{dx}{dy} - 2x = e^{y}$ ?

y = Ce^{2x} - e^{x}

Correct answerYour answer

x = Ce^{2y} + e^{y}

Correct answerYour answer

x = Ce^{2y} - e^{y}

Correct answerYour answer

x = Ce^{-2y} - e^{y}

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AnswerC · $x = Ce^{2y} - e^{y}$

Linear in $x$ with $P_{1} = -2$ : the factor is $e^{-2y}$ , and $\frac{d}{dy}\left(xe^{-2y}\right) = e^{-y}$ .
$xe^{-2y} = -e^{-y} + C$ , so $x = Ce^{2y} - e^{y}$ .

Worth noting

$y = Ce^{2x} - e^{x}$ swaps the roles of $x$ and $y$ ; here $x$ is the unknown.

Question 108

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Assertion (A): The solution of $\frac{dy}{dx} + y = xe^{-x}$ with $y = 1$ at $x = 0$ is $y = \left(\frac{x^{2}}{2} + 1\right)e^{-x}$ .

Reason (R): The integrating factor of $\frac{dy}{dx} + y = xe^{-x}$ is $e^{-x}$ .

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AnswerC · A is true, but R is false.

$P = 1$ , so the factor is $e^{x}$ and R is false: $ye^{x} = \int x\,dx = \frac{x^{2}}{2} + C$ .
$y(0) = C = 1$ , so $y = \left(\frac{x^{2}}{2} + 1\right)e^{-x}$ and A is true.

Worth noting

The factor comes from $P = 1$ , not from $Q = xe^{-x}$ , so it is $e^{x}$ .

Question 109

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For $x > 0$ , what is the general solution of $x\frac{dy}{dx} - y = x^{2}e^{x}$ ?

y = e^{x} + Cx

Correct answerYour answer

y = xe^{x} + C

Correct answerYour answer

y = x^{2}e^{x} + Cx

Correct answerYour answer

y = xe^{x} + Cx

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AnswerD · $y = xe^{x} + Cx$

Divide by $x$ : $\frac{dy}{dx} - \frac{1}{x}y = xe^{x}$ , with factor $\frac{1}{x}$ .
$\frac{y}{x} = \int e^{x}\,dx = e^{x} + C$ , so $y = xe^{x} + Cx$ .

Worth noting

$e^{x} + Cx$ multiplies only the constant by $x$ at the end; $e^{x}$ must be multiplied too.

Question 110

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For $0 < x < \pi$ , what is the general solution of $\frac{dy}{dx} + y\cot x = \operatorname{cosec} x$ ?

y = (x + C)\operatorname{cosec} x

Correct answerYour answer

y = (x + C)\sin x

Correct answerYour answer

y = x + C\operatorname{cosec} x

Correct answerYour answer

y = x\operatorname{cosec} x + C

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AnswerA · $y = (x + C)\operatorname{cosec} x$

The factor is $e^{\int \cot x\,dx} = \sin x$ , and $\operatorname{cosec} x \cdot \sin x = 1$ .
$y\sin x = \int 1\,dx = x + C$ , so $y = (x + C)\operatorname{cosec} x$ .

Worth noting

$(x + C)\sin x$ multiplies by the factor $\sin x$ where it should divide by it.

Question 111

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What is the general solution of $\frac{dx}{dy} + x = y$ ?

y = x - 1 + Ce^{-x}

Correct answerYour answer

x = y + 1 + Ce^{-y}

Correct answerYour answer

x = y - 1 + Ce^{-y}

Correct answerYour answer

x = y - 1 + Ce^{y}

Correct answerYour answer

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AnswerC · $x = y - 1 + Ce^{-y}$

Linear in $x$ with $P = 1$ : factor $e^{y}$ , and $\frac{d}{dy}(xe^{y}) = ye^{y}$ .
$xe^{y} = (y - 1)e^{y} + C$ , so $x = y - 1 + Ce^{-y}$ .

Worth noting

$y = x - 1 + Ce^{-x}$ swaps the roles of $x$ and $y$ ; here $x$ is the unknown.

Question 112

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What is the general solution of $x\frac{dy}{dx} + 2y = x^{2}$ ?

y = \frac{x^{2}}{4} + \frac{C}{x^{2}}

Correct answerYour answer

y = \frac{x^{2}}{2} + \frac{C}{x^{2}}

Correct answerYour answer

y = \frac{x^{2}}{4} + Cx^{2}

Correct answerYour answer

y = \frac{x^{4}}{4} + C

Correct answerYour answer

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AnswerA · $y = \frac{x^{2}}{4} + \frac{C}{x^{2}}$

Divide by $x$ : $\frac{dy}{dx} + \frac{2}{x}y = x$ , with factor $x^{2}$ .
$x^{2}y = \int x^{3}\,dx = \frac{x^{4}}{4} + C$ , so $y = \frac{x^{2}}{4} + \frac{C}{x^{2}}$ .

Worth noting

$\frac{x^{4}}{4} + C$ is $x^{2}y$ ; it still has to be divided by $x^{2}$ .

Question 113

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$y$ satisfies $\frac{dy}{dx} + y = 2x$ and $y = 0$ when $x = 0$ . What is $y$ when $x = 1$ ?

0

Correct answerYour answer

\frac{2}{e}

Correct answerYour answer

2 - \frac{2}{e}

Correct answerYour answer

\frac{4}{e}

Correct answerYour answer

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AnswerB · $\frac{2}{e}$

The factor $e^{x}$ : $ye^{x} = \int 2xe^{x}\,dx = 2(x - 1)e^{x} + C$ , so $y = 2x - 2 + Ce^{-x}$ .
$y(0) = -2 + C = 0$ gives $C = 2$ ; at $x = 1$ , $y = 0 + \frac{2}{e} = \frac{2}{e}$ .

Worth noting

0 drops the $Ce^{-x}$ term, taking $y = 2x - 2$ alone.

Question 114

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What is the general solution of $\frac{dy}{dx} + y\tan x = \sec x$ ?

y = \sin x + C\sec x

Correct answerYour answer

y = \tan x + C\cos x

Correct answerYour answer

y = \cos x + C\sin x

Correct answerYour answer

y = \sin x + C\cos x

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AnswerD · $y = \sin x + C\cos x$

The factor is $e^{\int \tan x\,dx} = \sec x$ : $\frac{d}{dx}(y\sec x) = \sec^{2}x$ .
$y\sec x = \tan x + C$ , so $y = \sin x + C\cos x$ .

Worth noting

$\tan x + C\cos x$ divides only the constant by $\sec x$ , not $\tan x$ .

Question 115

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A curve passes through $(0, 3)$ , and its slope at every point $(x, y)$ is $y - 2x$ . What is $y$ on the curve when $x = 1$ ?

4 + e

Correct answerYour answer

3e

Correct answerYour answer

4 + 3e

Correct answerYour answer

5e

Correct answerYour answer

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AnswerA · $4 + e$

$\frac{dy}{dx} - y = -2x$ has the factor $e^{-x}$ : $ye^{-x} = \int -2xe^{-x}\,dx = 2(x + 1)e^{-x} + C$ .
So $y = 2x + 2 + Ce^{x}$ ; at $(0, 3)$ , $2 + C = 3$ , so $C = 1$ .
At $x = 1$ : $y = 2 + 2 + e = 4 + e$ .

Worth noting

$5e$ comes from $y = 2x - 2 + Ce^{x}$ , a sign slip in $\int -2xe^{-x}\,dx$ ; then $C = 5$ .

Question 116

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With the balance ₹0 at $t = 0$ , after how long does it reach ₹2,40,000?

\frac{\log 2}{20}

yearsCorrect answerYour answer

20\log 2

yearsCorrect answerYour answer20 yearsCorrect answerYour answer

20\log 3

yearsCorrect answerYour answer

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AnswerB · $20\log 2$ years

$2{,}40{,}000\left(e^{t/20} - 1\right) = 2{,}40{,}000$ gives $e^{t/20} = 2$ .
$\frac{t}{20} = \log 2$ , so $t = 20\log 2$ , about 13.9 years.

Worth noting

20 years counts the deposits alone; with interest the balance gets there sooner.

Question 117

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For $y > 0$ , $x$ satisfies $\frac{dx}{dy} + \frac{2x}{y} = 4y$ , and $x = 2$ when $y = 1$ . What is $x$ when $y = 2$ ?

4

Correct answerYour answer

\frac{17}{4}

Correct answerYour answer

\frac{9}{2}

Correct answerYour answer

5

Correct answerYour answer

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AnswerB · $\frac{17}{4}$

Linear in $x$ with $P_{1} = \frac{2}{y}$ : the factor is $y^{2}$ , and $xy^{2} = \int 4y^{3}\,dy = y^{4} + C$ .
At $y = 1$ : $2 = 1 + C$ , so $C = 1$ and $x = y^{2} + \frac{1}{y^{2}}$ .
At $y = 2$ : $x = 4 + \frac{1}{4} = \frac{17}{4}$ .

Worth noting

5 uses $x = y^{2} + C$ : the constant was not divided by the factor $y^{2}$ .

Question 118

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$y$ satisfies $\frac{dy}{dx} + 2y = e^{x}$ , and $y = 1$ when $x = 0$ . What is $y$ when $x = \log 2$ ?

\frac{5}{6}

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\frac{11}{12}

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2

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\frac{10}{3}

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AnswerA · $\frac{5}{6}$

The factor is $e^{2x}$ : $ye^{2x} = \int e^{3x}\,dx = \frac{e^{3x}}{3} + C$ , so $y = \frac{e^{x}}{3} + Ce^{-2x}$ .
$y(0) = \frac{1}{3} + C = 1$ gives $C = \frac{2}{3}$ .
At $x = \log 2$ : $\frac{2}{3} + \frac{2}{3} \cdot \frac{1}{4} = \frac{5}{6}$ .

Worth noting

$\frac{11}{12}$ takes $C = 1$ , the starting value of $y$ , forgetting the $\frac{1}{3}$ that $y(0)$ already contains.

Question 119

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A tank holds 20 litres of pure water. Brine with 3 g of salt per litre flows in at 2 litres a minute, and the well-mixed solution flows out at 1 litre a minute, so after $t$ minutes the tank holds $20 + t$ litres and the salt $Q$ (in grams) satisfies $\frac{dQ}{dt} = 6 - \frac{Q}{20 + t}$ . How much salt is in the tank after 20 minutes?

3 gCorrect answerYour answer

120 - \frac{120}{e}

gCorrect answerYour answer90 gCorrect answerYour answer120 gCorrect answerYour answer

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AnswerC · 90 g

$\frac{dQ}{dt} + \frac{1}{20 + t}Q = 6$ has the factor $e^{\log(20 + t)} = 20 + t$ .
$Q(20 + t) = \int 6(20 + t)\,dt = 120t + 3t^{2} + C$ ; $Q(0) = 0$ gives $C = 0$ .
At $t = 20$ : $Q = \frac{2400 + 1200}{40} = 90$ g.

Worth noting

$120 - \frac{120}{e}$ keeps the volume at 20 litres, solving $\frac{dQ}{dt} = 6 - \frac{Q}{20}$ ; the tank fills by a litre each minute.

Question 120

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What is the general solution of $\frac{dy}{dx} + y = \cos x$ ?

y = \frac{\sin x + \cos x}{2} + Ce^{-x}

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y = \frac{\sin x - \cos x}{2} + Ce^{-x}

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y = \frac{\sin x + \cos x}{2} + Ce^{x}

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y = \sin x + Ce^{-x}

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AnswerA · $y = \frac{\sin x + \cos x}{2} + Ce^{-x}$

The factor is $e^{x}$ , so $ye^{x} = \int e^{x}\cos x\,dx$ .
By parts twice, $\int e^{x}\cos x\,dx = \frac{e^{x}(\sin x + \cos x)}{2} + C$ .
Divide by $e^{x}$ : $y = \frac{\sin x + \cos x}{2} + Ce^{-x}$ .

Worth noting

$\frac{\sin x - \cos x}{2}$ slips a sign in the second integration by parts: $\int e^{x}\sin x\,dx$ comes in with a minus.

NTA's Mathematics / Applied Mathematics syllabus has three sections: Section A1, which every candidate answers, then Section B1 (Mathematics) or Section B2 (Applied Mathematics), of which a candidate answers one. Each section lists units and topics but no chapters, so the chapters here are ours. Where both B sections list a topic that Section A also lists, its questions sit once, in Section A.