Solving a homogeneous equation with y = vx, for x > 0, gives √(1 + v^2) = Cx. What is the solution in terms of x and y?

Question 35

EasyOriginal question

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

Correct answerYour answer

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

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer A

Correct answer: A

$\sqrt{1 + \frac{y^{2}}{x^{2}}} = \frac{\sqrt{x^{2} + y^{2}}}{x}$ 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 $\frac{\sqrt{x^{2} + y^{2}}}{x}$ .

Ten questions from Homogeneous & Linear Differential Equations, at your own pace.