Help solving ODE using separation of variables

[opticaltempest]

Hello,

Could someone please help me to simplify my solution to my ODE?

Here is the solution I get when I check it using Maple 10,

http://img524.imageshack.us/img524/415/ode2hx.jpg

Here are my steps:

$$\left( {1 + x^3 } \right)\frac{{dy}}{{dx}} - 3x^2 y = 0$$

$$\left( {1 + x^3 } \right)dy = 3x^2 ydx$$

$$\int {\frac{1}{y}} dy = \int {\frac{{3x^2 }}{{1 + x^3 }}} dx$$

Let $$u = 1 + x^3$$ then $$\frac{{du}}{3} = x^2 dx$$

$$\ln \left| y \right| = \frac{1}{3}\int {\frac{1}{u}} du$$

$$\ln \left| y \right| = \frac{1}{3}\ln \left| u \right| + C$$

$$\ln \left| y \right| = \frac{1}{3}\ln \left| {1 + x^3 } \right| + C$$


How do I simplify this down to match the answer in Maple 10?

 

[Tide]

HINTS:

$$e^{\ln x} = x$$

$$\ln x^p = p \ln x$$

 

[StatusX]

That 1/3 shouldn't be there. Once that's gone, exp both sides to get the same expression as maple.