# Help solving ODE using separation of variables

1. May 9, 2006

### opticaltempest

Hello,

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

http://img524.imageshack.us/img524/415/ode2hx.jpg [Broken]

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?

Last edited by a moderator: May 2, 2017
2. May 9, 2006

### Tide

HINTS:

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

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

3. May 9, 2006

### StatusX

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