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:

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

<br /> \left( {1 + x^3 } \right)dy = 3x^2 ydx<br />

<br /> \int {\frac{1}{y}} dy = \int {\frac{{3x^2 }}{{1 + x^3 }}} dx<br />

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

<br /> \ln \left| y \right| = \frac{1}{3}\int {\frac{1}{u}} du<br />

<br /> \ln \left| y \right| = \frac{1}{3}\ln \left| u \right| + C<br />

<br /> \ln \left| y \right| = \frac{1}{3}\ln \left| {1 + x^3 } \right| + C<br />


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.