How to find the solution of the differential equation 
Find the solution of the differential equation 
  $$\frac{dy}{dx}=-\frac{x(x^2+y^2-10)}{y(x^2+y^2+5)}, y(0)=1$$

Trial: $$\begin{align} \frac{dy}{dx}=-\frac{x(x^2+y^2-10)}{y(x^2+y^2+5)} \\ \implies  \frac{dy}{dx}=-\frac{1+(y/x)^2-10/x^2}{(y/x)(1+(y/x)^2+5/x^2)} \\  \implies   v+x\frac{dv}{dx}=-\frac{1+v^2-10/x^2}{v(1+v^2+5/x^2)}  \end{align}$$ I can't seperate $x$ and $v$.

Your differential equation can be written as
$$ (x^{3}+x y^{2}-10 x)dx+(x^{2}y+y^{3}+5y)dy=0$$
So it is of the form $M(x,y)dx+N(x,y)dy=0$
Now, $$M_{y}=2xy$$ and $$N_{x}=2xy.$$
$\therefore$ The given differential equation is exact.
So its general solution is 
$$\int M dx+\int (\text{Terms in $N$ which does not contain $x$})dy=C$$
$$\Rightarrow \int (x^{3}+x y^{2}-10 x)dx+\int 0 dy=C$$
$$\Rightarrow \frac{x^{4}}{4}+\frac{x^{2}y^{2}}{2}-5x^{2}=c$$
Now $y(0)=1$, we have $C=0$.
$\therefore$ The solution of given differential equation is 
$\displaystyle \frac{x^{4}}{4}+\frac{x^{2}y^{2}}{2}-5x^{2}=0$.
