Differential equation. $ \sqrt{3+y^2}dx-x\cdot dy=x^2\cdot dy $ Check my steps

Question

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?

@KaviRamaMurthy like this? $\int \frac{dx}{-x(x+1)}$ and $-\int \frac{dy}{\sqrt{3+y^2}}$ — NewAtC, Oct 11 '21 at 08:30
In fact you don't have a minus sign on LHS, other thant that, yes. — zwim, Oct 11 '21 at 08:32
Your steps are correct and you no need of minus signs on both sides, you can cancel them and reciprocal on both sides. After integrate it which gives you the result. — RAHUL, Oct 11 '21 at 08:49

score 1 · Answer 1 · answered Oct 11 '21 at 09:12

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:

$$ y = \sinh\left(\log \dfrac{|x|}{|x+1|}-C\right). $$

1 Answers1