What is the general solution of (x - y)dy/dx = x + y?

tan^-1y/x = log√(x^2 + y^2) + C

tan^-1x/y = log√(x^2 + y^2) + C

tan^-1y/x = -log√(x^2 + y^2) + C

Question 55

HardOriginal question

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

Show the answer and explanation

CorrectNot quiteAnswer A

Correct answer: A

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

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