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Math Help - IVP -variation of parameters

  1. #1
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    IVP -variation of parameters

    ok so here is my *big* problem im stuck on..=[

    thank you for your time reading this

    dx/dt + x/t - 1 = t^2 ; x(1) = 1

    using variation of paramter i did all this:

    dx/dt + x/t = t^2 + 1

    dx/dt+ x/t = 0 [solve homogeneous equation first]

    integrating by parts i get. ..

    -lnx = -lnt + c

    x = e^(lnt+c) = At


    then i find x(t), x'(t)...

    x(t) = At, x'(t) = A

    now what i did is put these into the ode which was dx/dt + x/t - 1 = t^2

    and i get..

    A + At/t = t^2 + 1

    2A = t^2 + 1

    A = (t^2 + 1)/2

    and put A into x(t)...

    i get..

    x = ((t^2+1)/2)*t

    and x(1) = 1.... now im stuck!


    please could anyone guide me through or tell me what i did wrong? =[ am having a nightmare with these
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  2. #2
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    Quote Originally Posted by matlabnoob View Post
    ok so here is my *big* problem im stuck on..=[

    thank you for your time reading this

    dx/dt + x/t - 1 = t^2 ; x(1) = 1

    using variation of paramter i did all this:

    dx/dt + x/t = t^2 + 1

    dx/dt+ x/t = 0 [solve homogeneous equation first]

    integrating by parts i get. ..

    -lnx = -lnt + c

    x = e^(lnt+c) = At


    then i find x(t), x'(t)...

    x(t) = At, x'(t) = A

    now what i did is put these into the ode which was dx/dt + x/t - 1 = t^2

    and i get..

    A + At/t = t^2 + 1

    2A = t^2 + 1

    A = (t^2 + 1)/2

    and put A into x(t)...

    i get..

    x = ((t^2+1)/2)*t

    and x(1) = 1.... now im stuck!


    please could anyone guide me through or tell me what i did wrong? =[ am having a nightmare with these
    First off, the solution x = At doesn't solve the complementary ODE x' + \frac{x}{t} = 0 (it's x = \frac{A}{t} ). Second, for a variation of parameters let A = A(t), sub into the ODE and then reduce to an ODE for A(t).
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