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Math Help - Variation of parameter - 2nd order ODE

  1. #1
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    Thumbs down Variation of parameter - 2nd order ODE

    Find the general solution of this differential equation, given that a solution of the corresponding homogeneous equation is y = x,

    (x^2)y'' + x(x-2)y' - (x-2)y = x^3

    I can sorta deal with these problems but i havnt came across them with functions of x attached to the y's, I can also deal with them in the case of y'' + y' + y = f(x) :s... do i still find an auxillary equation or what? Can someone show me how these type of questions are done please? Thanks
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  2. #2
    MHF Contributor chisigma's Avatar
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    Let's start with the homogeneous equation...

    x^{2}\cdot y^{''} + x\cdot (x-2)\cdot y^{'} - (x-2)\cdot y = 0 (1)

    ... one solution of which we know to be y=x. If u(x) and v(x) are solutions of (1) then...

    u^{''} + (1-\frac{2}{x})\cdot u^{'} - \frac{x-2}{x^{2}}\cdot u=0

    v^{''} + (1-\frac{2}{x})\cdot v^{'} - \frac{x-2}{x^{2}}\cdot v=0 (2)

    Multipling the first equation by v, the second by u an taking the difference we obtain...

    (u^{''}\cdot v - u\cdot v^{''}) + (1-\frac{2}{x})\cdot (u^{'}\cdot v - u\cdot v^{'}) =0 (3)

    ... that with the substitution \gamma= u^{'}\cdot v - u\cdot v^{'} becomes...

    \gamma^{'} + (1-\frac{2}{x})\cdot \gamma =0 (4)

    The (4) is easily solved and gives...

    \gamma= c\cdot x^{2}\cdot e^{-x} (5)

    where c is a constant that can be set =1 without losing anything. Deviding both terms of (5) by v^{2} we obtain the new equation...

    \frac {u^{'}\cdot v - u\cdot v^{'}}{v^{2}}= \frac{d}{dx}(\frac{u}{v})= \frac{x^{2}}{v^{2}}\cdot e^{-x} (6)

    Since y=x is solution we can set v(x)=x in (6) obtaining...

    \frac {d}{dx}(\frac{u}{x})= e^{-x} (7)

    ... the solution of which is...

    u(x)= x\cdot e^{-x} (8)

    The general solution of (1) is then...

    y(x)= c_{1}\cdot x + c_{2}\cdot x\cdot e^{-x} (9)

    Let's consider now the 'complete' equation...


    x^{2}\cdot y^{''} + x\cdot (x-2)\cdot y^{'} - (x-2)\cdot y = x^{3} (10)

    With a little of patience you find that y=x^{2} is a particular solution so that the general solution of (10) is...

    y(x)= c_{1}\cdot x + c_{2}\cdot x\cdot e^{-x} + x^{2} (11)

    Kind regards

    \chi \sigma
    Last edited by chisigma; April 6th 2009 at 11:59 AM.
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  3. #3
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    Thanks for the work, but I dont really get what's going on though... I havn't been taught this at university yet however.

    What I'm familiar with is the following:

    ay'' + by' + cy = 0, where i obtain a general solution with either complex, repeated or real distinct

    ay'' + by' + cy = f(x)

    And I assume that the next installment of this would be the variation of parameters... i.e. what i've posted. Thanks again for the working, but i don't understand how ya got there.
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