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Math Help - Inhomogeneous equation

  1. #1
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    Inhomogeneous equation

    How can I solve this equation x''-x=(1+e^{t})^{-1} using the method of variation of parameters?Using the auxiliary equation the roots are r_{1}=0 and  r_{2}=1.
    I don't know if I'm suppose to look for a solution of the form x(t)=p+q*e^{x} manly because the initial equation has o 1+... after the equal sign .Can someone please help me?
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  2. #2
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    Quote Originally Posted by AkilMAI View Post
    How can I solve this equation x''-x=(1+e^{t})^{-1} using the method of variation of parameters?Using the auxiliary equation the roots are r_{1}=0 and  r_{2}=1.
    I don't know if I'm suppose to look for a solution of the form x(t)=p+q*e^{x} manly because the initial equation has o 1+... after the equal sign .Can someone please help me?
    I'm not clear on how well you know the method, so I'll just say this for now. Recheck your solution to the homogeneous equation. m^2 - 1 = 0 gives what as solutions?

    -Dan
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    m1=1 and m2=-1....I don't know why I had the impression that the equation is x''-x'=... and not x''-x=...
    I know that this method is called variation of parameters manly because we have varied the usual parameters
    ,A and B for instance, and made them functions x(t)=u1(t)x1(t)+u2(t)x2(t)
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  4. #4
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    but still I'm having trouble picking a solution to solve this
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by AkilMAI View Post
    m1=1 and m2=-1....I don't know why I had the impression that the equation is x''-x'=... and not x''-x=...
    I know that this method is called variation of parameters manly because we have varied the usual parameters
    ,A and B for instance, and made them functions x(t)=u1(t)x1(t)+u2(t)x2(t)
    Quote Originally Posted by AkilMAI View Post
    but still I'm having trouble picking a solution to solve this
    Since m = 1 and m = - 1 you know that the homogeneous solutions are
    x_1(t) = e^t~\text{ and }~x_2(t) = e^{-t}

    So your proposed particular solution will be
    x_p(t) = u_1(t)e^t + u_2(t)e^{-t}

    Can you work from there?

    -Dan
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