separable equation3

Question

Is this solution correct?

Equation:

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

solve:

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

Answer 1

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): $$C\exp\left(-\frac{1}{2}\ln(x^2+1)\right)= \sin y$$ $$y = \arcsin\left(C\exp\left(-\frac{1}{2}\ln(x^2+1)\right)\right)$$