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Thread: Reduction of Order

  1. #1
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    Reduction of Order

    $\displaystyle t^2y'' - 4ty' + 6y$

    $\displaystyle y1(t) = t^2
    $
    given one solution find another solution. I solved this using reduction of order.

    I got

    $\displaystyle v''t^4 = 0$

    where do I proceed from here tho?

    I know I have to integrate it or something but im forgetting the steps.
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  2. #2
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    Quote Originally Posted by p00ndawg View Post
    $\displaystyle t^2y'' - 4ty' + 6y$

    $\displaystyle y1(t) = t^2
    $
    given one solution find another solution. I solved this using reduction of order.

    I got

    $\displaystyle v''t^4 = 0$

    where do I proceed from here tho?

    I know I have to integrate it or something but im forgetting the steps.
    I assume the above equaiton is equal to zero....

    So the 2nd solution is of the form...

    $\displaystyle y=t^2\cdot u(t)$

    $\displaystyle y'=2tu+t^2u'$

    $\displaystyle y''=2u+2tu'+2tu'+t^2u''=t^2u''+4tu'+2u$

    Plugging in we get

    $\displaystyle t^2(t^2u''+4tu'+2u)-4t(2tu+t^2u')+6(t^2u)=0$

    $\displaystyle t^4u''+4t^3u'+2t^2u-8t^2u-4t^3u'+6(t^2u)=0$

    $\displaystyle t^4u''=0$

    So this implies that $\displaystyle u''=0$

    Now integrating twice we get

    $\displaystyle u'=c$ and $\displaystyle u=ct+d$ so if we choose c=1 and d=0 we get

    $\displaystyle u=t$

    so the 2nd solution is

    $\displaystyle
    y_2=c_2t^3
    $
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