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Help solving ODE using separation of variables

  1. May 9, 2006 #1
    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,

    [​IMG]

    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?
     
  2. jcsd
  3. May 9, 2006 #2

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    HINTS:

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

    [tex]\ln x^p = p \ln x[/tex]
     
  4. May 9, 2006 #3

    StatusX

    User Avatar
    Homework Helper

    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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