Differential Equation Initial Value Problem Here is a pretty standard initial value problem that I'm having a little trouble with.
$$(\ln(y))^2\frac{\mathrm{d}y}{\mathrm{d}x}=x^2y$$
Given $y(1)=e^2$, find the constant $C$.
So I separated and integrated to get $\frac{(\ln(y))^3}{3}=\frac{x^3}{3}+C$. Multiplying $3$ to both sides yields $(ln(y))^3=x^3+C$. Here's where I struggled a little.  So I take the cube root of both sides to get $\ln y=\sqrt[3]{x^3+C}$ then raised $e$ to both sides to get $y=e^\sqrt[3]{x^3+C}$...? Basically, the $C$ and the $e$ are giving me trouble.  I appreciate any pointers.

First of all, the $C$ in your first equation is not the $C$ in your second equation.  (If the first $C$ was called $C_1$, the second $C$ would be equal to $3C_1$.)
You have $y(1)=e^2$.  Plugging $x=1$ and $y=e^2$ into your last equation yields $e^2=e^\sqrt[3]{1^2+C}$, thus, $2=\sqrt[3]{1+C}$.  Can you do the rest?
