<s>[INST] How to find differential equation $$\frac{dy}{dx}-8x=2xy^2\quad  y=0\,x=1$$
I separated $x$ and $y$.
\begin{align*}
\color{red}{\frac{dy}{y^2}}&=\color{red}{2x+8x dx}\\
\frac{dy}{y^2}&=\color{red}{10x dx}\\
\color{red}{\ln y^2} &= 5x^2\\
y^2&=Ae^{5x^2}
\end{align*}
When I plug $y$ and $x$ in, i get $A=0$.
I think I did it wrong somewhere.
After I fixed it, I get.
$dy/y^2+4=2xdx$
$\ln(y^2+4)=x^2$
$y^2+4=Ae^{x^2}$
After plug in $y$ and $x$
$4=Ae^1$
$A=1.471$
 [/INST] $y′=2xy^2+8x⇒y′=x(2y^2+8).$ By separation of variables you find that $$y(x)=2\tan(2(2c_1+x^2))$$ and with the boundary condition $y(1 )=0$ the solution is $$y(x)=2\tan(2(x^2-1))$$ 
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