# solving :$\frac{dy}{dx} =-\frac{2}{y}-\frac{3y}{2x}$

How to solve : $$\frac{dy}{dx} =-\frac{2}{y}-\frac{3y}{2x}$$

-
Hint: It is seperable – N3buchadnezzar Oct 26 '12 at 8:52
It is not seperable, homo. and exact. – Babak S. Oct 26 '12 at 9:15

This equation can be exactly treated in this way. Assume $y\ne 0$ and multiply both sides by $y$. You will get $$y\frac{dy}{dx}+\frac{3}{2x}y^2==\frac{d}{dx}\frac{y^2}{2}+\frac{3}{x}\frac{y^2}{2}=-2.$$ Here just put $z=\frac{y^2}{2}$ and the equation becomes $$\frac{dz}{dx}+\frac{3}{x}z=-2$$ and this is very easy to solve. Write the solution as $$z=z_0(x)+kx$$ being $\frac{dz_0}{dx}+\frac{3}{x}z_0=0$ and, by substitution, $k=-\frac{1}{2}$. Finally, $z_0(x)=\frac{C}{x^3}$, being $C$ a constant. Finally, the solution can be written as $$\frac{y^2}{2}=\frac{C}{x^3}-\frac{1}{2}x.$$