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Non linear nonhomogenous ODE

  1. Nov 3, 2008 #1
    1. The problem statement, all variables and given/known data

    y' + y = t^2 , y(0) = 6, y'(0)= -6

    2. Relevant equations

    3. The attempt at a solution

    first i tried to seperate variables using y = ux but cant forward on and then i tried undetermined coeff. method. i found homogenous and particular solution but i am not sure about the solution because there is no need to use y'(0)=-6 and i really dont sure can i use this method for the first order ode.
  2. jcsd
  3. Nov 3, 2008 #2


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    If you show me your homogeneous and particular solutions, I'll stand a better chance of telling you what is wrong with them than if I just wildly guess at what you may have done wrong! ;0)
  4. Nov 3, 2008 #3
    x+1 = 0
    x = -1

    Yh = c1 e^-x

    Yp= K2 x^2 + K1x + K0 and take Yp' and write the Yp and Yp' to the eq'n find K2, K1 and K0

    Yp = x^2 - 2x +2

    Y = c1 e^-x + x^2 - 2x + 2

    there is only c1 and i dont need the second initial value
  5. Nov 3, 2008 #4


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    Shouldn't your y's be functions of t instead of x?!

    and since y'(t)+y(t)=t^2, y'(0)+y(0)=0 => y'(0)=-y(0) which is consistent with your initial values, and so you can use either of them to find c1.

    If on the other hand, you were given y(0)=6 and y'(0)=3, then there would be no solution since these initial conditions are inconsistent with your ODE.

    Luckily, your initial conditions are consistent and so your method and solution are correct! :smile:
  6. Nov 3, 2008 #5
    thanks for the help yes u r right it should be t . i dont like t :D
  7. Nov 3, 2008 #6


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    t is good for you! :biggrin:

    welcome to PF! :smile:
  8. Nov 3, 2008 #7
    thx :D but it confuses my mind :D

    by the way any other ways to solve this eq'n?
  9. Nov 3, 2008 #8


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    try "t" with ginseng....that should help your mind out :wink:

    ....as for other methods, I'm sure there are a few (such as power series solutions) but aside from plugging it into mathematica; this is the easiest way I know of.
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