Solve differential equations $y' = \frac{x-y^2}{2y(x+y^2)}$ Help me!
I'm can try solve this:
$$y' = \frac{x-y^2}{2y(x+y^2)}$$
I'm try to change variable
$$ t = y^2$$
but what is next?
am I need to solve this?
$$t'= \frac{x-t}{x+t}$$
 A: The equation is homogeneous so susbtitute $t=wx, t'=w'x+w$ You get a separable equation. Or you can solve it with exactness :
$$t'= \frac{x-t}{x+t}$$
$$(x+t)dt=(x-t)dx$$
$$xdt+tdx=xdx-tdt$$
Note that $xdt+tdx=dxt$
$$dxt=xdx-tdt$$
Integrate both sides
$$xt=\frac 1 2 (x^2- t^2)+C$$
Substitute back $t=y^2$
$$K=y^4 - x^2+2y^2x$$

You can also solve the original equation with exactness but without substitution:
$$y' = \frac{x-y^2}{2y(x+y^2)}$$
$${2y(x+y^2)}dy = ({x-y^2})dx$$
$$2yxdy+y^2dx=-2y^3dy +xdx$$
Note that $dy^2x=2yxdy+y^2dx$
$$dy^2x=-2y^3dy +xdx$$
Integrate:
$$C=y^4 - x^2+2y^2x$$
A: $$ 2y(x+y^2)dy=(x-y^2)dx $$ $\Rightarrow $
$$ xdx-y^2dx-2yxdy-2y^3dy=0$$ We can make a replacement (1)
$$ 1=2\alpha=\alpha+1+\alpha-1=3\alpha+\alpha-1 $$
$ \alpha=\frac{1}{2}$ $\Rightarrow $ $ y=\sqrt{z}$
$$\frac{dy}{dx}=\frac{dz}{2\sqrt{z}dz} $$ $\Rightarrow $
$$(x-z)dx-(x+z)dz=0 $$
$$\frac{x-z}{x+z}=\frac{dz}{dx} $$
Replacement (2)
$ \frac{z}{x}=m $ $\Rightarrow$ $$ \frac{dz}{dx}=\frac{dm}{dx}x+m$$
$$ \frac{1-m}{1+m}-m=\frac{dm}{dx}x$$
$$C+\int\frac{dx}{x}=\int\frac{dm}{\frac{1-m}{1+m}-m} $$
But we know that $$ \int\frac{(1+m)dm}{-m^2-2m+1}=\frac{-1}{2}\int\frac{d(m^2+2m-1)}{m^2+2m-1}=\frac{-1}{2}log(m^2+2m-1)$$ $$\Rightarrow $$
$$ \frac{C}{x^2}=m^2+2m-1$$ $m=\frac{y^2}{x} $
Final answer:
$$ C=y^4+2y^2x-x^2$$
A: A bit late answer but I think it is worth mentioning it.
I would suggest the following substitution:
$$t' =\frac{x-t}{x+t}\stackrel{u=t+x}{\Longrightarrow}u' = \frac{2x}{u}$$
Now, you get immediately
$$u(x) = \pm \sqrt{2}\sqrt{x^2+c}$$
What remains is backwards substitution $y^2=t=u-x$:
$$y(x)=\pm \sqrt{\pm \sqrt{2}\sqrt{x^2+c} - x}$$
