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Differential Equations for CUET Maths

About 5 min to learn · 67 questions

Differential equations are part of Section A of CUET Maths, the section every candidate answers, whether you take Mathematics or Applied Mathematics. CUET asks about them in a few fixed ways, so a little practice goes a long way. Learn each rule here, from order and degree to general and particular solutions, forming an equation from a family of curves, and solving by separating the variables, with growth and decay, 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 A1
On this page · 10 sections
  1. 01What a differential equation is
  2. 02Order and degree
  3. 03General and particular solutions
  4. 04Forming a differential equation
  5. 05Variables separable
  6. 06How CUET asks this chapter
  7. 07Exam shortcuts
  8. 08Common mistakes
  9. 09Quick revision
  10. 10Check your understanding

What a differential equation is

A dish holds 500 bacteria, and the colony grows at a rate proportional to its size: the more bacteria, the faster it grows. Written down, that sentence is dPdt=kP\frac{dP}{dt} = kP. It does not say what PP is; it says how PP changes. Finding PP from it is solving a differential equation, and by the end of the chapter we will know how many bacteria there are after 9 hours.

What is a differential equation?

A differential equation is an equation that contains a derivative of an unknown function, such as dydx=2x\frac{dy}{dx} = 2x or d2ydx2+y=0\frac{d^2y}{dx^2} + y = 0. Its solution is not a number but a function that makes the equation true for every value of the variable.

Order and degree

What are the order and degree of a differential equation?

The order is the order of the highest derivative in the equation. The degree is the power of that highest derivative, once the equation has been written as a polynomial in its derivatives, free of roots and fractional powers. If a derivative sits inside a function such as sin⁡\sin, log⁡\log or e  e^{\;}, the equation is not a polynomial in it and the degree is not defined.

Example 1 · Order and degree

Find the order and degree of (d2ydx2)3+(dydx)4+y=0\left(\frac{d^2y}{dx^2}\right)^3 + \left(\frac{dy}{dx}\right)^4 + y = 0.

  1. The highest derivative is d2ydx2\frac{d^2y}{dx^2}, so the order is 2.

  2. Its power is 3, so the degree is 3. The power 4 belongs to a lower derivative and does not count.

Answer

Order 2, degree 3.

Watch out

Clear roots and fractional powers before reading the degree. d2ydx2=1+dydx\frac{d^2y}{dx^2} = \sqrt{1 + \frac{dy}{dx}} squares to (d2ydx2)2=1+dydx\left(\frac{d^2y}{dx^2}\right)^2 = 1 + \frac{dy}{dx}, so its degree is 2, not 1. And d2ydx2+edydx=0\frac{d^2y}{dx^2} + e^{\frac{dy}{dx}} = 0 has order 2 but no degree.

General and particular solutions

What is the difference between a general and a particular solution?

The general solution contains as many arbitrary constants as the order of the equation, and describes a whole family of curves; y=x2+Cy = x^2 + C is the general solution of dydx=2x\frac{dy}{dx} = 2x. A particular solution has every constant fixed, usually by a given condition such as y=1y = 1 when x=0x = 0, which picks out y=x2+1y = x^2 + 1.

To check that a function is a solution, find its derivatives, substitute, and see whether the equation holds for every xx.

Example 2 · Checking a solution

Show that y=e−3xy = e^{-3x} is a solution of d2ydx2+dydx−6y=0\frac{d^2y}{dx^2} + \frac{dy}{dx} - 6y = 0.

  1. dydx=−3e−3x\frac{dy}{dx} = -3e^{-3x} and d2ydx2=9e−3x\frac{d^2y}{dx^2} = 9e^{-3x}.

  2. Substituting: 9e−3x−3e−3x−6e−3x=09e^{-3x} - 3e^{-3x} - 6e^{-3x} = 0.

Answer

The equation holds for every xx, so y=e−3xy = e^{-3x} is a solution.

Forming a differential equation

How do you form a differential equation from a family of curves?

Differentiate the family's equation as many times as it has arbitrary constants, then eliminate the constants between the equations you have. The result has order equal to the number of constants removed, which is why the equation of all straight lines, y=mx+cy = mx + c, has order 2.

Example 3 · Circles about the origin

Form the differential equation of the circles x2+y2=r2x^2 + y^2 = r^2, rr arbitrary.

  1. There is one constant, so differentiate once: 2x+2ydydx=02x + 2y\frac{dy}{dx} = 0.

  2. The constant rr has already disappeared.

Answer

x+ydydx=0x + y\frac{dy}{dx} = 0, of order 1.

Key idea

Two families come up so often that their equations are worth knowing. y=Acos⁡kx+Bsin⁡kxy = A\cos kx + B\sin kx gives d2ydx2+k2y=0\frac{d^2y}{dx^2} + k^2y = 0, and y=Aekxy = Ae^{kx} gives dydx=ky\frac{dy}{dx} = ky.

Variables separable

How do you solve a differential equation by separating the variables?

If the equation can be written as dydx=f(x)g(y)\frac{dy}{dx} = f(x)g(y), move every yy to the side of dydy and every xx to the side of dxdx: dyg(y)=f(x) dx\frac{dy}{g(y)} = f(x)\,dx. Then integrate both sides and add one constant.

Example 4 · Separating and integrating

Solve dydx=1+yx\frac{dy}{dx} = \frac{1 + y}{x}.

  1. Separate: dy1+y=dxx\frac{dy}{1 + y} = \frac{dx}{x}.

  2. Integrate: log⁡∣1+y∣=log⁡∣x∣+log⁡∣C∣\log|1 + y| = \log|x| + \log|C|, writing the constant as a logarithm.

Answer
1+y=Cx1 + y = Cx

Example 5 · A particular solution

Solve dydx=3x2y\frac{dy}{dx} = 3x^2y, given y=2y = 2 when x=0x = 0.

  1. dyy=3x2 dx\frac{dy}{y} = 3x^2\,dx, so log⁡∣y∣=x3+c\log|y| = x^3 + c and y=Cex3y = Ce^{x^3}.

  2. At x=0x = 0: 2=Ce02 = Ce^0, so C=2C = 2.

Answer
y=2ex3y = 2e^{x^3}

Back to the bacteria. dPP=k dt\frac{dP}{P} = k\,dt gives P=P0ektP = P_0e^{kt}, with P0=500P_0 = 500 at the start. Growth proportional to size means the colony multiplies by the same factor in equal times. If it doubles every 3 hours, then after 9 hours it has doubled three times: 500×23=4000500 \times 2^3 = 4000 bacteria. Decay works the same way with dNdt=−kN\frac{dN}{dt} = -kN: each half-life halves what is left.

Watch out

Growth and decay multiply; they never add. A colony that doubles every 3 hours does not gain 500 every 3 hours, and a sample that halves every 5 years never reaches 0.

How CUET asks this chapter

  • Order and degree: of a given equation, their sum or product, or a statement set, with roots, fractional powers and "not defined" as the traps.
  • Solutions: which function solves an equation, how many constants a general solution has, or a particular solution from a condition.
  • Forming: the equation of a family such as y=Ae3xy = Ae^{3x} or y=Asin⁡x+Bcos⁡xy = A\sin x + B\cos x, or the order of the equation of all circles or all lines.
  • Separable equations: a general solution, the value of yy at a point from a condition, or growth and decay.

Exam shortcuts

  • Order = number of arbitrary constants in the family.
  • All lines: order 2; circles about the origin: order 1; all circles: order 3.
  • dydx=ky\frac{dy}{dx} = ky has solution y=Cekxy = Ce^{kx}.
  • Doubling time TT: P=P0×2t/TP = P_0 \times 2^{t/T}; half-life TT: N=N0×(12)t/TN = N_0 \times \left(\frac{1}{2}\right)^{t/T}.
  • Check an answer by differentiating it and substituting.

Common mistakes

  • Reading the degree from a lower derivative's power.
  • Reading the degree before clearing a root or a fractional power.
  • Giving a degree when a derivative sits inside sin⁡\sin, log⁡\log or e  e^{\;}.
  • Forgetting the constant, or adding a separate constant on each side.
  • Treating growth or decay as adding a fixed amount.

Quick revision

Order and degree

  • Order: highest derivative; degree: its power
  • Clear roots first; inside sin⁡\sin or e  e^{\;}: degree not defined

Solutions

  • General: as many constants as the order; particular: none left
  • Forming: differentiate once per constant, then eliminate

Separable

  • dyg(y)=f(x) dx\frac{dy}{g(y)} = f(x)\,dx, integrate, one constant
  • dPdt=kP\frac{dP}{dt} = kP: P=P0ektP = P_0e^{kt}

The last two chapters of Section A turn from calculus to chance and to choices: probability.

Next · 09Quick revision

Check your understanding

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

Order and degreeQ1 of 67Q1

Order and degree

15 questions

Question 1

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What is the order of d2ydx2+3dydx−4y=0\frac{d^2y}{dx^2} + 3\frac{dy}{dx} - 4y = 0?

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AnswerB · 22

  1. The order is the order of the highest derivative in the equation.

  2. The highest derivative here is d2ydx2\frac{d^2y}{dx^2}, so the order is 2.

Question 2

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What is the degree of (dydx)3+2y=x\left(\frac{dy}{dx}\right)^3 + 2y = x?

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AnswerD · 3

  1. The highest derivative is dydx\frac{dy}{dx}, of order 1.

  2. Its highest power is 3, so the degree is 3.

Worth noting

1 is the order; the degree is the power on that highest derivative.

Question 3

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What is the order of d3ydx3+2d2ydx2−dydx=x\frac{d^3y}{dx^3} + 2\frac{d^2y}{dx^2} - \frac{dy}{dx} = x?

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AnswerC · 33

  1. The order is the order of the highest derivative in the equation.

  2. The highest derivative here is d3ydx3\frac{d^3y}{dx^3}, so the order is 3.

Worth noting

1 is the degree, the power of the highest derivative; the order counts how many times yy is differentiated.

Question 4

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What is the degree of d2ydx2=(1+(dydx)2)2\frac{d^2y}{dx^2} = \left(1 + \left(\frac{dy}{dx}\right)^2\right)^2?

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AnswerC · 1

  1. The highest derivative is d2ydx2\frac{d^2y}{dx^2}, and it appears to the power 1.

  2. Expanding the bracket gives (dydx)4\left(\frac{dy}{dx}\right)^4, but the power on a lower derivative does not count: the degree is 1.

Worth noting

4 is the power of dydx\frac{dy}{dx}, the lower derivative.

Question 5

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What is the degree of d2ydx2+sin⁡(dydx)=0\frac{d^2y}{dx^2} + \sin\left(\frac{dy}{dx}\right) = 0?

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AnswerD · Not defined

  1. The degree is defined only when the equation is a polynomial in its derivatives.

  2. Here a derivative sits inside sin⁡\sin, which is not a polynomial, so the degree is not defined.

Worth noting

The order is still defined: it is 2, from the highest derivative.

Question 6

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What is the product of the order and the degree of (d3ydx3)2+(d2ydx2)3+dydx=0\left(\frac{d^3y}{dx^3}\right)^2 + \left(\frac{d^2y}{dx^2}\right)^3 + \frac{dy}{dx} = 0?

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AnswerC · 66

  1. The highest derivative is d3ydx3\frac{d^3y}{dx^3}: the order is 3.

  2. Its power is 2, so the degree is 2 (the cube on d2ydx2\frac{d^2y}{dx^2} does not count), and the product is 6.

Worth noting

9 takes the degree as 3, the power on d2ydx2\frac{d^2y}{dx^2}, a lower derivative.

Question 7

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Which of the following statements are true?

I. The order of a differential equation is always a positive integer.

II. The degree of a differential equation is always defined.

III. dydx=cos⁡x\frac{dy}{dx} = \cos x has order 1 and degree 1.

IV. Squaring a derivative changes the order of the equation.

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AnswerB · I and III only

  1. The order counts derivatives, so it is a positive integer (I); and dydx=cos⁡x\frac{dy}{dx} = \cos x is a polynomial of degree 1 in its only derivative (III).

  2. II is false: with a derivative inside sin⁡\sin or e  e^{\;}, the degree is not defined. IV is false: a power changes the degree, not the order.

Question 8

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What is the order of (dydx)4−3xd2ydx2=0\left(\frac{dy}{dx}\right)^4 - 3x\frac{d^2y}{dx^2} = 0?

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AnswerA · 22

  1. The order is the order of the highest derivative in the equation.

  2. The highest derivative here is d2ydx2\frac{d^2y}{dx^2}, so the order is 2.

Worth noting

A power on a derivative changes the degree, never the order: (dydx)4\left(\frac{dy}{dx}\right)^4 is still a first derivative.

Question 9

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Which of these differential equations has order 2 and degree 1?

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AnswerD · d2ydx2+y=0\frac{d^2y}{dx^2} + y = 0

  1. Order 2 needs d2ydx2\frac{d^2y}{dx^2} as the highest derivative; degree 1 needs it to the power 1.

  2. Only d2ydx2+y=0\frac{d^2y}{dx^2} + y = 0 has both. The others have order 1 and degree 2, order 2 and degree 2, and order 3.

Worth noting

(dydx)2+y=0\left(\frac{dy}{dx}\right)^2 + y = 0 swaps the two: order 1, degree 2.

Question 10

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What is the degree of dydx+edydx=x\frac{dy}{dx} + e^{\frac{dy}{dx}} = x?

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AnswerD · Not defined

  1. The degree is defined only when the equation is a polynomial in its derivatives.

  2. edydxe^{\frac{dy}{dx}} is not a polynomial in dydx\frac{dy}{dx}, so the degree is not defined (the order is 1).

Worth noting

1 is the order; an exponential of a derivative leaves the degree undefined.

Question 11

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What is the sum of the order and the degree of (d3ydx3)2+d2ydx2+(dydx)5=0\left(\frac{d^3y}{dx^3}\right)^2 + \frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^5 = 0?

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AnswerA · 55

  1. The highest derivative is d3ydx3\frac{d^3y}{dx^3}: order 3.

  2. It appears squared, so the degree is 2, and 3+2=53 + 2 = 5.

Worth noting

8 takes the degree from (dydx)5\left(\frac{dy}{dx}\right)^5, the highest power in sight, not the power of the highest derivative.

Question 12

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What is the degree of (1+(dydx)2)32=d2ydx2\left(1 + \left(\frac{dy}{dx}\right)^2\right)^{\frac{3}{2}} = \frac{d^2y}{dx^2}?

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AnswerB · 2

  1. First clear the fractional power by squaring: (1+(dydx)2)3=(d2ydx2)2\left(1 + \left(\frac{dy}{dx}\right)^2\right)^3 = \left(\frac{d^2y}{dx^2}\right)^2.

  2. Now the highest derivative, d2ydx2\frac{d^2y}{dx^2}, is to the power 2: the degree is 2.

Worth noting

1 reads the degree before the fractional power is cleared.

Question 13

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What is the degree of y=xdydx+1+(dydx)2y = x\frac{dy}{dx} + \sqrt{1 + \left(\frac{dy}{dx}\right)^2}?

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AnswerA · 2

  1. Clear the root: (y−xdydx)2=1+(dydx)2\left(y - x\frac{dy}{dx}\right)^2 = 1 + \left(\frac{dy}{dx}\right)^2.

  2. Expanding, (x2−1)(dydx)2−2xydydx+y2−1=0(x^2 - 1)\left(\frac{dy}{dx}\right)^2 - 2xy\frac{dy}{dx} + y^2 - 1 = 0: the only derivative is dydx\frac{dy}{dx}, to the power 2, so the degree is 2.

Worth noting

1 reads the degree before the root is removed.

Question 14

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Which of the following statements are true?

I. The order of d2ydx2+(dydx)3=0\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^3 = 0 is 2.

II. The degree of d2ydx2=dydx+1\sqrt{\frac{d^2y}{dx^2}} = \frac{dy}{dx} + 1 is 1.

III. The degree of d2ydx2+cos⁡(dydx)=0\frac{d^2y}{dx^2} + \cos\left(\frac{dy}{dx}\right) = 0 is 1.

IV. When the degree is defined, it is a positive integer.

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AnswerA · I, II and IV only

  1. I: the highest derivative is d2ydx2\frac{d^2y}{dx^2}. II: squaring gives d2ydx2=(dydx+1)2\frac{d^2y}{dx^2} = \left(\frac{dy}{dx} + 1\right)^2, where d2ydx2\frac{d^2y}{dx^2} has power 1. IV: the degree is the power of a derivative in a polynomial.

  2. III is false: cos⁡(dydx)\cos\left(\frac{dy}{dx}\right) is not a polynomial in dydx\frac{dy}{dx}, so the degree is not defined.

Question 15

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What is the degree of d2ydx2=1+dydx\frac{d^2y}{dx^2} = \sqrt{1 + \frac{dy}{dx}}?

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AnswerB · 2

  1. Clear the root by squaring: (d2ydx2)2=1+dydx\left(\frac{d^2y}{dx^2}\right)^2 = 1 + \frac{dy}{dx}.

  2. The highest derivative, d2ydx2\frac{d^2y}{dx^2}, is now squared: the degree is 2.

Worth noting

1 reads the degree before the root is cleared.

General and particular solutions

15 questions

Question 16

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Which of these is a solution of d2ydx2−y=0\frac{d^2y}{dx^2} - y = 0?

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AnswerD · y=exy = e^{x}

  1. Try y=exy = e^{x}: y′=exy' = e^{x} and y′′=exy'' = e^{x}.

  2. Substituting makes both sides equal for every xx; each of the others leaves something over.

Question 17

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How many arbitrary constants does the general solution of a differential equation of order 3 contain?

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AnswerD · 33

  1. Solving an equation of order nn takes nn integrations, and each brings one constant.

  2. So the general solution of an order-3 equation has 3 arbitrary constants.

Worth noting

0 is the number in a particular solution, where every constant has been given a value.

Question 18

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What is the general solution of d2ydx2=0\frac{d^2y}{dx^2} = 0?

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AnswerA · y=ax+by = ax + b

  1. Integrating once, dydx=a\frac{dy}{dx} = a; integrating again, y=ax+by = ax + b.

  2. A second-order equation needs two arbitrary constants, aa and bb: every straight line.

Worth noting

y=axy = ax has one constant, too few for a second-order equation; it is only a particular family.

Question 19

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Which is the particular solution of dydx=2x\frac{dy}{dx} = 2x with y=3y = 3 when x=0x = 0?

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AnswerC · y=x2+3y = x^2 + 3

  1. The general solution is y=x2+Cy = x^2 + C.

  2. y(0)=3y(0) = 3 gives C=3C = 3, so the particular solution is y=x2+3y = x^2 + 3.

Worth noting

y=x2+Cy = x^2 + C is the general solution; the condition is what picks one curve from it.

Question 20

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Which of these is a solution of xdydx=yx\frac{dy}{dx} = y?

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AnswerB · y=3xy = 3 x

  1. Try y=3xy = 3 x: y′=3y' = 3.

  2. Substituting makes both sides equal for every xx; each of the others leaves something over.

Question 21

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Read the two statements and choose the correct option.

I. y=Aexy = Ae^{x}, AA arbitrary, is the general solution of dydx=y\frac{dy}{dx} = y.

II. y=2exy = 2e^{x} is a particular solution of dydx=y\frac{dy}{dx} = y.

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AnswerC · Both I and II

  1. I: ddx(Aex)=Aex=y\frac{d}{dx}(Ae^{x}) = Ae^{x} = y for every AA, and an order-1 equation needs one constant. True.

  2. II: A=2A = 2 is one choice of the constant, so it is a particular solution. True.

Question 22

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Which of these is a solution of d2ydx2+4y=0\frac{d^2y}{dx^2} + 4y = 0?

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AnswerC · y=sin⁡2xy = \sin 2x

  1. Try y=sin⁡2xy = \sin 2x: y′=2cos⁡2xy' = 2 \cos 2x and y′′=−4sin⁡2xy'' = - 4 \sin 2x.

  2. Substituting makes both sides equal for every xx; each of the others leaves something over.

Question 23

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Assertion (A): y=x2+Cy = x^2 + C is a solution of dydx=2x\frac{dy}{dx} = 2x for every constant CC.

Reason (R): The derivative of a constant is 0.

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

  1. ddx(x2+C)=2x+0=2x\frac{d}{dx}(x^2 + C) = 2x + 0 = 2x, so A is true.

  2. It works for every CC precisely because the constant's derivative is 0, so R explains A.

Question 24

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Which is the particular solution of dydx=cos⁡x\frac{dy}{dx} = \cos x with y=2y = 2 when x=0x = 0?

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AnswerA · y=sin⁡x+2y = \sin x + 2

  1. Integrating, y=sin⁡x+Cy = \sin x + C.

  2. At x=0x = 0: 2=0+C2 = 0 + C, so C=2C = 2 and y=sin⁡x+2y = \sin x + 2.

Worth noting

y=sin⁡x+Cy = \sin x + C is the general solution; the condition must fix CC. y=cos⁡x+1y = \cos x + 1 passes through (0,2)(0, 2) but has slope −sin⁡x-\sin x.

Question 25

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Which of these is a solution of dydx=2xy\frac{dy}{dx} = 2xy?

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AnswerA · y=ex2y = e^{x^{2}}

  1. Try y=ex2y = e^{x^{2}}: y′=2xex2y' = 2 x e^{x^{2}}.

  2. Substituting makes both sides equal for every xx; each of the others leaves something over.

Question 26

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Assertion (A): The general solution of d2ydx2=0\frac{d^2y}{dx^2} = 0 has two arbitrary constants.

Reason (R): The number of arbitrary constants in a general solution equals the degree of the differential equation.

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AnswerC · A is true, but R is false.

  1. Integrating twice gives y=ax+by = ax + b, with two constants: A is true.

  2. R is false: the number of constants equals the order. d2ydx2=0\frac{d^2y}{dx^2} = 0 has order 2 but degree 1.

Question 27

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Which of these is a solution of xdydx+y=0x\frac{dy}{dx} + y = 0?

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AnswerC · y=1xy = \frac{1}{x}

  1. Try y=1xy = \frac{1}{x}: y′=−1x2y' = - \frac{1}{x^{2}}.

  2. Substituting makes both sides equal for every xx; each of the others leaves something over.

Question 28

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d2ydx2=6x\frac{d^2y}{dx^2} = 6x, with y=1y = 1 and dydx=2\frac{dy}{dx} = 2 when x=0x = 0. What is yy?

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AnswerC · y=x3+2x+1y = x^3 + 2x + 1

  1. Integrate once: dydx=3x2+A\frac{dy}{dx} = 3x^2 + A; dydx=2\frac{dy}{dx} = 2 at x=0x = 0 gives A=2A = 2.

  2. Integrate again: y=x3+2x+By = x^3 + 2x + B; y=1y = 1 at x=0x = 0 gives B=1B = 1.

Worth noting

y=x3+1y = x^3 + 1 uses only the condition on yy; an order-2 equation needs both constants fixed.

Question 29

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Which of these is a solution of dydx+y=0\frac{dy}{dx} + y = 0?

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AnswerD · y=5e−xy = 5 e^{- x}

  1. Try y=5e−xy = 5 e^{- x}: y′=−5e−xy' = - 5 e^{- x}.

  2. Substituting makes both sides equal for every xx; each of the others leaves something over.

Question 30

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Which of these is a solution of d2ydx2−2dydx+y=0\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 0?

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AnswerB · y=xexy = x e^{x}

  1. Try y=xexy = x e^{x}: y′=xex+exy' = x e^{x} + e^{x} and y′′=(x+2)exy'' = \left(x + 2\right) e^{x}.

  2. Substituting makes both sides equal for every xx; each of the others leaves something over.

Forming a differential equation

17 questions

Question 31

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Which differential equation does the family y=Ae3xy = Ae^{3x} satisfy, AA being arbitrary constants?

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AnswerB · dydx=3y\frac{dy}{dx} = 3y

  1. There is one constant, so differentiate once and eliminate it.

  2. Eliminating gives dydx=3y\frac{dy}{dx} = 3y, of order 1.

Worth noting

The order of the equation equals the number of arbitrary constants it removes.

Question 32

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Which differential equation does the family y=mxy = mx satisfy, mm being arbitrary constants?

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AnswerA · xdydx=yx\frac{dy}{dx} = y

  1. There is one constant, so differentiate once and eliminate it.

  2. Eliminating gives xdydx=yx\frac{dy}{dx} = y, of order 1.

Worth noting

The order of the equation equals the number of arbitrary constants it removes.

Question 33

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What is the order of the differential equation of all circles with centre at the origin?

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AnswerA · 11

  1. The order equals the number of independent arbitrary constants in the family's equation.

  2. x2+y2=r2x^2 + y^2 = r^2 has one constant, rr: the order is 1.

Question 34

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Which differential equation does the family y=Ax2y = Ax^2 satisfy, AA being an arbitrary constant?

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AnswerB · xdydx=2yx\frac{dy}{dx} = 2y

  1. dydx=2Ax\frac{dy}{dx} = 2Ax, and from the family A=yx2A = \frac{y}{x^2}.

  2. So dydx=2yx\frac{dy}{dx} = \frac{2y}{x}, that is xdydx=2yx\frac{dy}{dx} = 2y.

Worth noting

dydx=2x\frac{dy}{dx} = 2x sets A=1A = 1 instead of eliminating AA: it describes one parabola, not the family.

Question 35

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What is the order of the differential equation of the family y=Aex+Be2x+Ce3xy = Ae^{x} + Be^{2x} + Ce^{3x}, with A,B,CA, B, C arbitrary?

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AnswerC · 33

  1. Each arbitrary constant takes one differentiation to remove.

  2. Three constants, so the order is 3.

Worth noting

6 adds the powers 1+2+31 + 2 + 3; only the number of constants counts.

Question 36

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Which differential equation does the family y=Asin⁡x+Bcos⁡xy = A\sin x + B\cos x satisfy, A,BA, B being arbitrary constants?

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AnswerD · d2ydx2+y=0\frac{d^2y}{dx^2} + y = 0

  1. There are two constants, so differentiate twice and eliminate them.

  2. Eliminating gives d2ydx2+y=0\frac{d^2y}{dx^2} + y = 0, of order 2.

Worth noting

The order of the equation equals the number of arbitrary constants it removes.

Question 37

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Which differential equation does the family y2=4axy^2 = 4ax satisfy, aa being arbitrary constants?

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AnswerD · y=2xdydxy = 2x\frac{dy}{dx}

  1. There is one constant, so differentiate once and eliminate it.

  2. Eliminating gives y=2xdydxy = 2x\frac{dy}{dx}, of order 1.

Worth noting

The order of the equation equals the number of arbitrary constants it removes.

Question 38

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Which family of curves has the differential equation dydx=−2y\frac{dy}{dx} = -2y?

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AnswerD · y=Ae−2xy = Ae^{-2x}

  1. For y=Ae−2xy = Ae^{-2x}, dydx=−2Ae−2x=−2y\frac{dy}{dx} = -2Ae^{-2x} = -2y, for every AA.

  2. Ae2xAe^{2x} gives +2y+2y, and −2x+A-2x + A gives −2-2, not −2y-2y.

Worth noting

y=Ax−2y = Ax^{-2} gives dydx=−2yx\frac{dy}{dx} = -\frac{2y}{x}, with an extra xx.

Question 39

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What is the order of the differential equation of all straight lines in the plane?

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AnswerC · 22

  1. The order equals the number of independent arbitrary constants in the family's equation.

  2. y=mx+cy = mx + c has two constants, mm and cc: the order is 2.

Question 40

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Which of the following statements are true?

I. The differential equation of y=Aexy = Ae^{x} has order 1.

II. Eliminating 2 arbitrary constants gives an equation of order 2.

III. The differential equation of y=Aex+Be−xy = Ae^{x} + Be^{-x} is d2ydx2=y\frac{d^2y}{dx^2} = y.

IV. The differential equation of the lines y=mx+cy = mx + c is dydx=m\frac{dy}{dx} = m.

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AnswerD · I, II and III only

  1. I and II: one differentiation per constant. III: d2ydx2=Aex+Be−x=y\frac{d^2y}{dx^2} = Ae^{x} + Be^{-x} = y.

  2. IV is false: mm is one of the constants and must be removed too; the equation is d2ydx2=0\frac{d^2y}{dx^2} = 0.

Question 41

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Which differential equation does the family y=Ax+By = \frac{A}{x} + B satisfy, A,BA, B being arbitrary constants?

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AnswerD · xd2ydx2+2dydx=0x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = 0

  1. dydx=−Ax2\frac{dy}{dx} = -\frac{A}{x^2}, so x2dydx=−Ax^2\frac{dy}{dx} = -A.

  2. Differentiate again: 2xdydx+x2d2ydx2=02x\frac{dy}{dx} + x^2\frac{d^2y}{dx^2} = 0, that is xd2ydx2+2dydx=0x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = 0.

Worth noting

xdydx+y=0x\frac{dy}{dx} + y = 0 is the equation of y=Axy = \frac{A}{x} alone; the +B+B needs a second derivative to remove.

Question 42

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Which family of curves has the differential equation dydx=yx\frac{dy}{dx} = \frac{y}{x}?

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AnswerA · Straight lines through the origin, y=Cxy = Cx

  1. Separate: dyy=dxx\frac{dy}{y} = \frac{dx}{x}, so log⁡∣y∣=log⁡∣x∣+log⁡∣C∣\log|y| = \log|x| + \log|C|.

  2. That is y=Cxy = Cx: every line through the origin.

Worth noting

xy=Cxy = C has dydx=−yx\frac{dy}{dx} = -\frac{y}{x}, the same equation with a minus sign.

Question 43

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What is the order of the differential equation of all parabolas y2=4a(x−h)y^2 = 4a(x - h), whose axis is the xx-axis, aa and hh arbitrary?

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AnswerB · 22

  1. The family has two arbitrary constants, aa and hh.

  2. Removing both takes two differentiations: the order is 2.

Worth noting

1 counts only aa; the vertex (h,0)(h, 0) moves too, so hh is a second constant.

Question 44

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Which differential equation does the family y=Aex+Be−2xy = Ae^{x} + Be^{-2x} satisfy, A,BA, B being arbitrary constants?

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AnswerB · d2ydx2+dydx−2y=0\frac{d^2y}{dx^2} + \frac{dy}{dx} - 2y = 0

  1. There are two constants, so differentiate twice and eliminate them.

  2. Eliminating gives d2ydx2+dydx−2y=0\frac{d^2y}{dx^2} + \frac{dy}{dx} - 2y = 0, of order 2.

Worth noting

The order of the equation equals the number of arbitrary constants it removes.

Question 45

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What is the order of the differential equation of all circles in the plane?

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AnswerB · 33

  1. The order equals the number of independent arbitrary constants in the family's equation.

  2. (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2 has three constants, hh, kk and rr: the order is 3.

Question 46

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Which differential equation describes the circles x2+y2=2axx^2 + y^2 = 2ax, with centre on the xx-axis and passing through the origin?

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AnswerB · 2xydydx=y2−x22xy\frac{dy}{dx} = y^2 - x^2

  1. Differentiate: 2x+2ydydx=2a2x + 2y\frac{dy}{dx} = 2a, so a=x+ydydxa = x + y\frac{dy}{dx}.

  2. Put this back: x2+y2=2x(x+ydydx)x^2 + y^2 = 2x\left(x + y\frac{dy}{dx}\right), which gives 2xydydx=y2−x22xy\frac{dy}{dx} = y^2 - x^2.

Worth noting

ydydx=a−xy\frac{dy}{dx} = a - x still contains the constant aa; a differential equation of the family must not.

Question 47

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Which differential equation does the family y=ex(A+Bx)y = e^{x}(A + Bx) satisfy, A,BA, B being arbitrary constants?

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AnswerD · d2ydx2−2dydx+y=0\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 0

  1. ye−x=A+Bxye^{-x} = A + Bx. Differentiate twice to remove AA, then BB: d2dx2(ye−x)=0\frac{d^2}{dx^2}\left(ye^{-x}\right) = 0.

  2. That is e−x(d2ydx2−2dydx+y)=0e^{-x}\left(\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y\right) = 0, so d2ydx2−2dydx+y=0\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 0.

Worth noting

d2ydx2−y=0\frac{d^2y}{dx^2} - y = 0 is the equation of Aex+Be−xAe^{x} + Be^{-x}; the repeated exe^{x} gives −2dydx-2\frac{dy}{dx}.

Variables separable

20 questions

Question 48

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

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AnswerA · y2−x2=Cy^2 - x^2 = C

  1. Separate the variables, with every yy on the side of dydy and every xx on the side of dxdx.

  2. Integrating both sides and collecting the constants gives y2−x2=Cy^2 - x^2 = C.

Worth noting

Check an answer by differentiating it: it must give back the equation.

Question 49

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Which equation has its variables separable?

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AnswerB · dydx=x2y\frac{dy}{dx} = x^2y

  1. Separable means dydx=f(x)g(y)\frac{dy}{dx} = f(x)g(y): a product of a function of xx alone and a function of yy alone.

  2. x2yx^2y is such a product, so dyy=x2 dx\frac{dy}{y} = x^2\,dx. A sum such as x+yx + y cannot be split this way.

Worth noting

x+yx + y is a sum, not a product; that equation is linear, a Section B1 method.

Question 50

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If dydx=yx\frac{dy}{dx} = \frac{y}{x} and y=2y = 2 when x=1x = 1, what is yy when x=3x = 3?

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AnswerC · 66

  1. Separating, log⁡∣y∣=log⁡∣x∣+log⁡∣C∣\log|y| = \log|x| + \log|C|, so y=Cxy = Cx.

  2. y(1)=2y(1) = 2 gives C=2C = 2: y=2xy = 2x, and y(3)=6y(3) = 6.

Worth noting

5 adds 3 to the starting value, treating the slope as 1.

Question 51

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What is the general solution of dydx=x2y2\frac{dy}{dx} = \frac{x^2}{y^2}?

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AnswerC · y3−x3=Cy^3 - x^3 = C

  1. Separate the variables, with every yy on the side of dydy and every xx on the side of dxdx.

  2. Integrating both sides and collecting the constants gives y3−x3=Cy^3 - x^3 = C.

Worth noting

Check an answer by differentiating it: it must give back the equation.

Question 52

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Which of these differential equations is NOT of the variables separable type?

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AnswerB · dydx=x+y\frac{dy}{dx} = x + y

  1. Separable means dydx=g(x)h(y)\frac{dy}{dx} = g(x)h(y), a product of a function of xx and a function of yy.

  2. xyxy, xy\frac{x}{y} and exeye^{x}e^{y} are such products; x+yx + y is a sum and cannot be split this way.

Worth noting

exey=ex+ye^{x}e^{y} = e^{x + y} looks like a sum in the power, but it is a product of two functions.

Question 53

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What is the general solution of dydx=ex−y\frac{dy}{dx} = e^{x - y}?

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AnswerC · ey=ex+Ce^y = e^x + C

  1. Separate the variables, with every yy on the side of dydy and every xx on the side of dxdx.

  2. Integrating both sides and collecting the constants gives ey=ex+Ce^y = e^x + C.

Worth noting

Check an answer by differentiating it: it must give back the equation.

Question 54

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What is the general solution of dydx=2xy\frac{dy}{dx} = 2xy?

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AnswerB · y=Cex2y = Ce^{x^2}

  1. Separate the variables, with every yy on the side of dydy and every xx on the side of dxdx.

  2. Integrating both sides and collecting the constants gives y=Cex2y = Ce^{x^2}.

Worth noting

Check an answer by differentiating it: it must give back the equation.

Question 55

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What is the general solution of dydx=y2\frac{dy}{dx} = y^2?

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AnswerA · −1y=x+C-\frac{1}{y} = x + C

  1. Separate the variables, with every yy on the side of dydy and every xx on the side of dxdx.

  2. Integrating both sides and collecting the constants gives −1y=x+C-\frac{1}{y} = x + C.

Worth noting

Check an answer by differentiating it: it must give back the equation.

Question 56

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If dydx=2xy\frac{dy}{dx} = 2xy and y=3y = 3 when x=0x = 0, what is yy when x=1x = 1?

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AnswerC · 3e3 e

  1. Separating and integrating gives the general solution; the condition y(0)=3y(0) = 3 fixes the constant.

  2. The particular solution is y=3ex2y = 3 e^{x^{2}}.

  3. At x=1x = 1: y=3ey = 3 e.

Worth noting

Leaving out the constant fixed by the condition gives the wrong curve of the family.

Question 57

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A population grows at a rate proportional to its size, dPdt=kP\frac{dP}{dt} = kP. It is 1,000 now and doubles in 5 years. What will it be after 15 years?

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AnswerD · 8,0008{,}000

  1. dPP=k dt\frac{dP}{P} = k\,dt gives P=P0ektP = P_0e^{kt}.

  2. Doubling in 5 years means e5k=2e^{5k} = 2, so P=1000×2t/5P = 1000 \times 2^{t/5}.

  3. At t=15t = 15: 1000×23=80001000 \times 2^3 = 8000.

Worth noting

6,000 adds 2,000 every 5 years, as if doubling were adding; growth multiplies.

Question 58

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A radioactive sample decays at a rate proportional to the amount present. It has 800 g now and a half-life of 5 years. How much is left after 15 years, in grams?

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AnswerC · 100100

  1. dNdt=−kN\frac{dN}{dt} = -kN gives N=N0e−ktN = N_0e^{-kt}, and each half-life halves what is left.

  2. 15 years is 3 half-lives: 800×(12)3=100800 \times \left(\frac{1}{2}\right)^3 = 100 g.

Worth noting

0 takes away 400 g each half-life, a straight line; decay halves what is left, so it never reaches 0.

Question 59

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What is the general solution of dydx=1+y2\frac{dy}{dx} = 1 + y^2?

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AnswerA · tan⁡−1y=x+C\tan^{-1} y = x + C

  1. Separate the variables, with every yy on the side of dydy and every xx on the side of dxdx.

  2. Integrating both sides and collecting the constants gives tan⁡−1y=x+C\tan^{-1} y = x + C.

Worth noting

Check an answer by differentiating it: it must give back the equation.

Question 60

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What is the general solution of dydx=ytan⁡x\frac{dy}{dx} = y\tan x?

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AnswerD · y=Csec⁡xy = C\sec x

  1. Separate the variables, with every yy on the side of dydy and every xx on the side of dxdx.

  2. Integrating both sides and collecting the constants gives y=Csec⁡xy = C\sec x.

Worth noting

Check an answer by differentiating it: it must give back the equation.

Question 61

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

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AnswerB · ex+e−y=Ce^x + e^{-y} = C

  1. Separate the variables, with every yy on the side of dydy and every xx on the side of dxdx.

  2. Integrating both sides and collecting the constants gives ex+e−y=Ce^x + e^{-y} = C.

Worth noting

Check an answer by differentiating it: it must give back the equation.

Question 62

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A bacteria culture grows at a rate proportional to its size and triples in 2 hours. By what factor has it grown after 6 hours?

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AnswerD · 2727

  1. dNdt=kN\frac{dN}{dt} = kN gives N=N0ektN = N_0e^{kt}, and e2k=3e^{2k} = 3.

  2. After 6 hours: e6k=(e2k)3=27e^{6k} = \left(e^{2k}\right)^3 = 27.

Worth noting

9 multiplies by 3 for 6 hours divided by 2 hours instead of raising 3 to that power.

Question 63

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If dydx=e−y\frac{dy}{dx} = e^{-y} and y=0y = 0 when x=0x = 0, what is yy when x=e−1x = e - 1?

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AnswerA · 11

  1. Separate: ey dy=dxe^{y}\,dy = dx, so ey=x+Ce^{y} = x + C; y(0)=0y(0) = 0 gives C=1C = 1.

  2. y=log⁡(x+1)y = \log(x + 1), and at x=e−1x = e - 1, y=log⁡e=1y = \log e = 1.

Worth noting

e−1e - 1 copies the value of xx; yy grows like a logarithm, much more slowly.

Question 64

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What is the general solution of dydx=1+y21+x2\frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}?

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AnswerD · tan⁡−1y=tan⁡−1x+C\tan^{-1} y = \tan^{-1} x + C

  1. Separate the variables, with every yy on the side of dydy and every xx on the side of dxdx.

  2. Integrating both sides and collecting the constants gives tan⁡−1y=tan⁡−1x+C\tan^{-1} y = \tan^{-1} x + C.

Worth noting

Check an answer by differentiating it: it must give back the equation.

Question 65

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If dydx=xy\frac{dy}{dx} = \frac{x}{y} and y=3y = 3 when x=0x = 0, what is yy when x=4x = 4?

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AnswerB · 55

  1. Separating and integrating gives the general solution; the condition y(0)=3y(0) = 3 fixes the constant.

  2. The particular solution is y=x2+9y = \sqrt{x^{2} + 9}.

  3. At x=4x = 4: y=5y = 5.

Worth noting

Leaving out the constant fixed by the condition gives the wrong curve of the family.

Question 66

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If dydx=ycos⁡x\frac{dy}{dx} = y\cos x and y=1y = 1 when x=0x = 0, what is yy when x=π2x = \frac{\pi}{2}?

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AnswerA · ee

  1. Separating and integrating gives the general solution; the condition y(0)=1y(0) = 1 fixes the constant.

  2. The particular solution is y=esin⁡xy = e^{\sin x}.

  3. At x=π2x = \frac{\pi}{2}: y=ey = e.

Worth noting

Leaving out the constant fixed by the condition gives the wrong curve of the family.

Question 67

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The slope of a curve at every point (x,y)(x, y) is 2yx\frac{2y}{x}, and the curve passes through (1,3)(1, 3). What is yy when x=2x = 2?

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AnswerC · 1212

  1. dyy=2 dxx\frac{dy}{y} = \frac{2\,dx}{x}, so log⁡∣y∣=2log⁡∣x∣+log⁡∣C∣\log|y| = 2\log|x| + \log|C|: y=Cx2y = Cx^2.

  2. Through (1,3)(1, 3): C=3C = 3. At x=2x = 2, y=12y = 12.

Worth noting

6 treats the curve as the line y=3xy = 3x; the factor 2 in the slope makes it a parabola.

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.