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Is this solution correct?

Equation:

$x \sin y~dx+(x^2+1)\cos y~dy=0$

solve:

$$x \sin y\,dx=-(x^2+1)\cos y\,dy \\ \frac{x}{-(x^2+1)}dx=\frac{\cos y}{\sin y}\,dy \\ \frac{x}{-(x^2+1)}dx=\cot y\,dy \\ \implies -\frac{1}{2}\ln(|x^2+1|)=-\cot^{-1} y \,dy$$.

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I would call this a separable equation, not "impartible", which gives the opposite idea. –  Pedro Tamaroff Mar 3 '13 at 19:10
    
Yes .soryy.Thank you for the reminder –  Software Mar 3 '13 at 19:28
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1 Answer

up vote 3 down vote accepted

You almost had it, but not quite.

From your second-to-last line: $$\frac{-x}{(x^2+1}\,dx = \cot(y)\,dy$$ $$\int\frac{-x}{(x^2+1)}\,dx = \int\cot(y)\,dy$$ $$-\frac{1}{2}\ln(x^2+1) + C= \int\frac{\cos(y)}{\sin(y)}\,dy$$ For the left integral, let $u = \sin(y) \implies du=\cos(y) \,dy$. $$-\frac{1}{2}\ln(x^2+1) + C= \int\frac{du}{u}\,dy$$ $$-\frac{1}{2}\ln(x^2+1) + C= \ln|\sin y|$$

This can be solved for $y$, but it's not pretty (and may introduce some domain error): $$C\exp\left(-\frac{1}{2}\ln(x^2+1)\right)= \sin y$$ $$y = \arcsin\left(C\exp\left(-\frac{1}{2}\ln(x^2+1)\right)\right)$$

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@flashdesign: You may vote up the answers you find helpful and accept the one that helped you the best. –  Gigili Mar 3 '13 at 18:57
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