Solve $4x^2y(x)y'(x)+y^4(x)+x^2=0$. I want solve $4x^2y(x)y'(x)+y^4(x)+x^2=0$.
I know that the way is make a substitution, but I don't find it... Any idea? Thanks
 A: $$4x^2y(x)y'(x)+y(x)^4+x^2=0\Longleftrightarrow$$

Let $v(x)=y(x)^2$:

$$4x^2\cdot\sqrt{v(x)}\cdot\frac{v'(x)}{2\sqrt{v(x)}}+v(x)^2+x^2=0\Longleftrightarrow$$
$$4x^2\cdot\frac{v'(x)}{2}+v(x)^2+x^2=0\Longleftrightarrow$$

Let $v(x)=xu(x)$ which gives $v'(x)=u(x)+xu'(x)$:

$$x^2+2x^2\left(xu'(x)+u(x)\right)+x^2u(x)^2=0\Longleftrightarrow$$
$$x^2\left(2xu'(x)+u(x)^2+2u(x)+1\right)=0\Longleftrightarrow$$
$$u'(x)=\frac{-u(x)^2-2u(x)-1}{2x}\Longleftrightarrow$$
$$\frac{2u'(x)}{-u(x)^2-2u(x)-1}=\frac{1}{x}\Longleftrightarrow$$
$$\int\frac{2u'(x)}{-u(x)^2-2u(x)-1}\space\text{d}x=\int\frac{1}{x}\space\text{d}x\Longleftrightarrow$$
$$\frac{2}{u(x)+1}=\ln|x|+\text{C}\Longleftrightarrow$$
$$u(x)=-\frac{\ln|x|+\text{C}-2}{\ln|x|+\text{C}}\Longleftrightarrow$$
$$v(x)=-\frac{x\left(\ln|x|+\text{C}-2\right)}{\ln|x|+\text{C}}\Longleftrightarrow$$
$$y(x)=\pm\sqrt{-\frac{x\left(\ln|x|+\text{C}-2\right)}{\ln|x|+\text{C}}}\Longleftrightarrow$$
$$y(x)=\pm\sqrt{\frac{x\left(\ln|x|-2\right)-2x\text{C}}{2\text{C}-\ln|x|}}$$
