I have to solve this initial value problem and determine where the solution attains its minimum value. This is the differential equation : $y'= 4y^2 + xy^2 , y(0)=1.$
I was able to find the solution $y$ for this equation which is :
$$y=\frac{-2}{8 x + (x^2 - 2)},$$
but I don't know how to determine where the solution attains its minimum value.
So any help with it would be appreciated.
 A: You can take the derivative of your solution (which is correct, by the way), and set it equal to zero to find the maximum value. Or you can work from the original DE:
\begin{align*}
y'&=0\\
4y^2+xy^2&=0\\
y^2(4+x)&=0.
\end{align*}
One solution is $y=0,$ and the other is $x=-4.$ But in looking at your solution, $y\not=0$ holds everywhere! So the only solution is $x=-4.$ But this doesn't tell us whether we have a max or a min here. Let us try the second derivative test:
\begin{align*}
y'&=y^2(4+x)\\
y''&=2yy'(4+x)+y^2\\
&=2yy^2(4+x)(4+x)+y^2\\
&=2y^3(4+x)^2+y^2.
\end{align*}
Now, at $x=-4,$ we have that $y''=y^2\ge 0,$ whereas the solution tells us that
$$y(-4)=\frac{-2}{-32 + (16 - 2)}=\frac{-2}{-18}=\frac19>0. $$
This tells us that $x=-4$ is a relative min, which is not what you are after. 
So, moving on: are there any other critical points? Yes! Where the function $y$ blows up, which occurs at
\begin{align*}
x^2 +8x - 2&=0\\
x&=\frac{-8\pm\sqrt{64-4(-2)}}{2}\\
&=\frac{-8\pm\sqrt{72}}{2}\\
&=\frac{-8\pm 6\sqrt{2}}{2}\\
&=-4\pm 3\sqrt{2}.
\end{align*}
$y$ is not defined here, so these values are not in the domain. We conclude that there is no absolute maximum!
