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Math Help - 2nd ODEs

  1. #1
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    2nd ODEs

    Having a few problems

    1. d^2y/dx^2 + 4dy/dx + 5y= x + 2

    So the characteristic equation is r^2+4r+5=0
    with imaginary roots so

    y(h)= e^-2x(C(1)cosx+ C(2)sinx)

    But how do i solve for y(p)? I cant do 5y=x+2.. thats not working

    I think i'll be fine for all of them once I figure out how to solve these when there is an x variable on the right side and not just a 0 or a constant
    Last edited by Jeffman50; April 25th 2010 at 11:53 AM.
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  2. #2
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    The solution to an inhomogenous linear equation is the sum of the general solution to the homogenous equation plus a particular solution. Its not easy in general to find a particular solution; most people just guess and check I think. In your case, try guessing y_{\text{particular}}(x) = bx+c and solve for b and c. Then the general solution will be y =e^{-2x}(C_1\cos x+ C_2\sin x) + bx + c.
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  3. #3
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    Are there any strategies other than just guessing?
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  4. #4
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    In general? Not that I am aware of. If the RHS is a polynomial and the LHS has constant coefficients, the particular soluiton will always be a polynomial of the same degree however.
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  5. #5
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    Quote Originally Posted by Jeffman50 View Post
    Are there any strategies other than just guessing?
    http://www.mathhelpforum.com/math-he...-tutorial.html

    See posts #6,7,8
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