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.

$$y'(x)\ln^2(y(x))=y(x)x^2\Longleftrightarrow$$
$$\frac{y'(x)\ln^2(y(x))}{y(x)}=x^2\Longleftrightarrow$$
$$\int\frac{y'(x)\ln^2(y(x))}{y(x)}\space\text{d}x=\int x^2\space\text{d}x\Longleftrightarrow$$

For the LHS, subtitute $u=\ln(y(x))$ and $\text{d}u=\frac{y'(x)}{y(x)}\space\text{d}x$.

$$\frac{\ln^3(y(x))}{3}=\frac{x^3}{3}+\text{C}$$
Now, use $y(1)=e^2$:
$$\frac{\ln^3(e^2)}{3}=\frac{1^3}{3}+\text{C}\Longleftrightarrow\text{C}=\frac{7}{3}$$
So, we get:
$$\ln^3(y(x))=7+x^3$$
Now, because of the initial condition, you can only use the real solution:
$$y(x)=\exp\left[\sqrt[3]{7+x^3}\right]$$
