Treating $x$ as the unknown, what is the integrating factor of $(1 + y^{2})\,dx = (2\tan^{-1}y - x)\,dy$ ?

e^{-\tan^{-1}y}

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e^{\tan^{-1}x}

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e^{\tan^{-1}y}

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1 + y^{2}

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Show the answer and explanation

CorrectNot quiteAnswer C

Correct answer: C

$\frac{dx}{dy} = \frac{2\tan^{-1}y - x}{1 + y^{2}}$ , so $\frac{dx}{dy} + \frac{x}{1 + y^{2}} = \frac{2\tan^{-1}y}{1 + y^{2}}$ .
$P_{1} = \frac{1}{1 + y^{2}}$ , so the factor is $e^{\int \frac{dy}{1 + y^{2}}} = e^{\tan^{-1}y}$ .

Worth noting

$e^{-\tan^{-1}y}$ keeps $-x$ on the right; moved to the left it becomes $+\frac{x}{1 + y^{2}}$ .

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

Treating xxx as the unknown, what is the integrating factor of (1+y2) dx=(2tan⁡−1y−x) dy(1 + y^{2})\,dx = (2\tan^{-1}y - x)\,dy(1+y2)dx=(2tan−1y−x)dy?