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Thread: Solving a 2nd order ODE with reduction of order.

  1. #1
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    Solving a 2nd order ODE with reduction of order.

    I need help on solving this. Our professor requires that we must use this format:

    W=u' and W'=u"

    Picture of it is down below.

    I do not know what to do once I have W'=0 my integrating factor would be weird


    http://oi67.tinypic.com/2rdf2w6.jpg
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    Re: Solving a 2nd order ODE with reduction of order.

    What's weird about it? It's about as straightforward as you are going to get.

    -Dan
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    Re: Solving a 2nd order ODE with reduction of order.

    Because W'=0. Therefore, that means my int. factor is e^(integral of 0) which = e^c.

    so then we times the integrating factor by the whole equation....

    then d/dx [e^c*W]=0
    integrate.

    W=c1*e^(-c)

    W=u'... we need to find u because y=uy1

    therefore...

    integral of W...=?

    I just feel like I am doing this wrong.


    The solution to y_2= xe^2x
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    Forum Admin topsquark's Avatar
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    Re: Solving a 2nd order ODE with reduction of order.

    Quote Originally Posted by math951 View Post
    Because W'=0. Therefore, that means my int. factor is e^(integral of 0) which = e^c.

    so then we times the integrating factor by the whole equation....

    then d/dx [e^c*W]=0
    integrate.

    W=c1*e^(-c)

    W=u'... we need to find u because y=uy1

    therefore...

    integral of W...=?

    I just feel like I am doing this wrong.


    The solution to y_2= xe^2x
    W' = 0 means that W = c = u' means that u = cx + C

    Thus $\displaystyle y = (cx + C)e^{2x}$. Now plug that into your original equation.

    -Dan
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    Re: Solving a 2nd order ODE with reduction of order.

    Thank you.
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