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Math Help - General solution of a Second order differential equations

  1. #1
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    General solution of a Second order differential equations

    Hi,

    Heres the question:

    Find the general solution of the following ordinary differential equation

    \frac{d^2y}{dx^2} + 4\frac{dy}{dx} + 5y = x

    So far I have found out the complementary function but I can't seem to work out the partial function.

    y_{pi} = kx,  \frac{dy_{pi}}{dx} = k , \frac{d^2y_{pi}}{dx^2} = 0

    so 5kx +4k = x

    How do i solve the above
    Last edited by Beard; March 3rd 2010 at 10:58 AM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    First we find the general integral of the 'incomplete equation' solving the second order equation...

    s^{2} + 4s + 5 = 0 (1)

    ... that has the solutions s= -2 \pm i. The general integral of the 'incomplete equation' is then...

    y_{g} (x) = e^{-2x}\cdot (c_{1}\cdot \cos x + c_{2}\cdot \sin x) (2)

    The next step is to find a particular integral of the 'complete equation'. At this purpose we search a function of the type y_{p} (x) = a\cdot x + b. If we substitute y_{p} (x) in the DE we find a= \frac{1}{5} and b= - \frac{4}{25}. The general integral of the 'complete equation' is then...

    y(x) = y_{g} (x) + y_{p} (x) = e^{-2x}\cdot (c_{1}\cdot \cos x + c_{2}\cdot \sin x) + \frac{1}{5}\cdot x -\frac{4}{25} (3)

    Kind regards

    \chi \sigma
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  3. #3
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    I got the general equation as the same but where did you manage to get the a= 1/5 and b = -4/25 from?
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  4. #4
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Beard View Post
    I got the general equation as the same but where did you manage to get the a= 1/5 and b = -4/25 from?
    The DE is...

    y^{''} + 4 y^{'} + 5 y = x (1)

    ... and we search a particular solution of the form...

    y = a\cdot x + b \rightarrow y^{'}= a , y^{''}=0 (2)

    Now subsituting (2) in (1) we obtain...

    5ax + 5b + 4a = x (3)

    ... so that is a = \frac{1}{5} , b = - \frac{4}{25}...

    Kind regards

    \chi \sigma
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  5. #5
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    EDIT: I'm an idiot
    Last edited by Beard; March 3rd 2010 at 02:16 PM.
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  6. #6
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    Quote Originally Posted by Beard View Post
    EDIT: I'm an idiot
    Well, I can't speak to that but it is generally true that when you have x^n on the right side, you will need to try a polynomial of the form Ax^n+ Bx^{n-1}+ \cdot\cdot\cdot+ Yx+ Z- that is all powers up to n.
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