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Math Help - Initial condition problem

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    Initial condition problem

    y''-2y'+2y=x+1, y(0) = 3, y'(0)=0

    I haven't done this in some time, please help.

    I know I need to find the homogenous solution and add it to a particular solution.

    Homogenous:

    r^2-2r+2 = 0

    So r = \{ 1+i , 1-i \} , then the homogenous solution is y_{h} = c_{1}e^{x}cosx+c_{2}e^xsinx

    Particular:

    Let y = Ax + B, y'=A, y''=0

    0 -2 (A) + 2(Ax+B) = x+1

    A=1/2, B=1, y_{p}= \frac {1}{2} x + 1

    yh + yp =  c_{1}e^{x}cosx+c_{2}e^xsinx +  \frac {1}{2} x + 1
    Last edited by tttcomrader; March 26th 2008 at 11:02 AM.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    y''-2y'+2y=x+1, y(0) = 3, y'(0)=0

    I haven't done this in some time, please help.

    I know I need to find the homogenous solution and add it to a particular solution.

    Homogenous:

    r^2-2r+2 = 0

    So r = \{ 1+i , 1-i \}
    You can do it the same way you do with real solutions:
    y(x) = Ae^{(1 + i)x} + Be^{(1 - i)x}

    We typically would want a real expression for this, so we may do the following:
    y(x) = e^x(Ae^{ix} + Be^{-ix})

    = e^x(Acos(x) + Ai~sin(x) + Bcos(x) - Bi~sin(x))

    = e^x((A + B)cos(x) + (A - B)i~sin(x))
    and we may replace the two arbitrary constants A and B with two new arbitrary constants:
    y(x) = e^x(a~cos(x) + b~sin(x))

    (What would happen here if you solved the problem using the original exponent form is that A and B would end up being complex constants.)

    So
    y(0) = 3 = a

    y^{\prime}(0) = a + b = 0

    So a = 3 and b = -3, giving
    y(x) = 3e^x(cos(x) - sin(x))

    I wasn't sure... Are you asking for help with the particular solution as well? Or do you have a handle on that?

    -Dan
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