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Math Help - Reduction of Order

  1. #1
    Newbie staevobr's Avatar
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    [SOLVED ] Reduction of Order

    x(1+2x)y'' + (1+6x)y' + 2y = \frac{1}{x^2}

    y_1=\frac{1}{1+2x} is a solution to do related homogeneous equation, and I need to find a second linearly independent solution of the DE using reduction of order.

    So, y_2 = V(x)\frac{1}{1+2x}

    y_2' = \frac{V'}{1+2x} - \frac{2V}{(1+2x)^2}

    y_2'' = \frac{V''}{1+2x} - \frac{4V'}{(1+2x)^2} + \frac{8V}{(1+2x)^3}

    Plugging y_2 into the homogeneous DE:

    x(1+2x)[\frac{V''}{1+2x} - \frac{4V'}{(1+2x)^2} + \frac{8V}{(1+2x)^3}] + (1+6x)[\frac{V'}{1+2x} - \frac{2V}{(1+2x)^2}] + 2[V(x)\frac{1}{1+2x}] = 0

    After some simplification, I get

    \frac{(4x^3+4x^2+x)V'' + (12x^3+3x)V' + (8x^2+8x)V}{(1+2x)^2} = 0

    I know that for reduction of order to work, the terms involving V need to go away, but I can't see a way to make that happen.
    Last edited by staevobr; October 17th 2010 at 09:08 PM.
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    why not make it generally at once?

    consider a(x)y''(x)+b(x)y'(x)+c(x)y=0, by knowing that y_1(x) is a solution of the homogeneous equation, then substitute y_2(x)=y_1(x)u and prove that u(x)=\displaystyle\int\frac1{y_1^2}e^{-\int\frac{b(x)}{a(x)}\,dx}\,dx.
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  3. #3
    Newbie staevobr's Avatar
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    My teacher doesn't allow the use of that formula.
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    completely useless, it's worth of time to prove that formula and then apply it.

    well, there's nothing for me to say here.
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  5. #5
    Newbie staevobr's Avatar
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    I found an error in my simplification.
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