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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?

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?

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