<s>[INST] 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?
$$t'= \frac{x-t}{x+t}$$
 [/INST] $$ 2y(x+y^2)dy=(x-y^2)dx $$ $\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:
$$ C=y^4+2y^2x-x^2$$
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