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Help solving ODE using separation of variables

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 [Broken]

Here are my steps:

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

[tex]
\left( {1 + x^3 } \right)dy = 3x^2 ydx
[/tex]

[tex]
\int {\frac{1}{y}} dy = \int {\frac{{3x^2 }}{{1 + x^3 }}} dx
[/tex]

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

[tex]
\ln \left| y \right| = \frac{1}{3}\int {\frac{1}{u}} du
[/tex]

[tex]
\ln \left| y \right| = \frac{1}{3}\ln \left| u \right| + C
[/tex]

[tex]
\ln \left| y \right| = \frac{1}{3}\ln \left| {1 + x^3 } \right| + C
[/tex]


How do I simplify this down to match the answer in Maple 10?
 
Last edited by a moderator:

Tide

Science Advisor
Homework Helper
3,077
0
HINTS:

[tex]e^{\ln x} = x[/tex]

[tex]\ln x^p = p \ln x[/tex]
 

StatusX

Homework Helper
2,563
1
That 1/3 shouldn't be there. Once that's gone, exp both sides to get the same expression as maple.
 

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