What is the general solution of dy/dx = y/x + sin^2y/x?

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

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

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

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CorrectNot quiteAnswer C

Correct answer: 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: $\int \operatorname{cosec}^{2}v\,dv = -\cot v$ .

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