I have equation $ \sqrt{3+y^2}dx-x\cdot dy=x^2\cdot dy $ and I need to solve it with separation method.
My try:
$-x^2\cdot dy-x\cdot dy=-\sqrt{3+y^2}\cdot dx$
$-x(x+1)\cdot dy=-\sqrt{3+y^2}\cdot dx $
$\frac{-x(x+1)}{dx}=-\frac{\sqrt{3+y^2}}{dy}$
And another step should be integrals for both sides. Are my steps correct or not?
Just rewrite the equation as $$ \dfrac{1}{x+x^2} dx = \frac{1}{\sqrt{3+y^2}}dy. $$
By integration, you get $$ \log \frac{|x|}{|1+x|} = \textrm{arcsinh} y + C $$
So, the solution can be given in an explicit form:
$$ y = \sinh\left(\log \dfrac{|x|}{|x+1|}-C\right). $$