For x > 0, what is the general solution of xdy/dx - y = 2√(x^2 + y^2)?

Question 59

HardOriginal question

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

Show the answer and explanation

CorrectNot quiteAnswer B

Correct answer: B

$y = vx$ : $x\frac{dv}{dx} = 2\sqrt{1 + v^{2}}$ , 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 $x$ .

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