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This should be a simple problem to solve but im just not getting it

  1. Oct 5, 2004 #1
    I'm given an equation to solve and I need to use the "method of undetermined coefficients" all I need to find is A,B,C,D I'm not given any initial values so I cant solve the entire thing. This is what I have but I'm not sure if it's correct or not. Thanks for any help...

    Equation: 2y''+5y'-3y=-2t^3+4t^2

    My work:
    2r^2+5r-3=0
    r=1/2, r=-3
    Yh=C1e^(1/2t) + C2e^(-3t)

    Assume Yp=(At^3+Bt^2+Ct+D)

    yp'=3At^2-2Bt+C
    yp''=6At-2B

    plugging the derivatives into the original equation I get: 12At-4B+15At^2-10Bt+5C-3At^3-3Bt^2-3Ct-3D=-2t^3+4t^2

    Solving for A I get: -3At^3=-2t^3 -> A=2/3

    Solving for B I get: 10t^2-3Bt^2=4t^2 -> B=2 ?

    Solving for C I get: 8t-20t-3Ct=0 -> C=-4 ?

    Solving for D I get:-8-20-3D=0 -> D=-28/3 ?

    The "?" are the ones I dont know are right. I'm pretty sure the value for A is right though. Also, I plugged in the values for A, B, C, D into the equation as I solved for them, incase you were wondering where some of the numbers came from...

    I really need help with this, so anything you can do will be greatly appreciated.
     
  2. jcsd
  3. Oct 5, 2004 #2

    arildno

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    "Assume Yp=(At^3+Bt^2+Ct+D)

    yp'=3At^2-2Bt+C
    yp''=6At-2B

    plugging the derivatives into the original equation I get: 12At-4B+15At^2-10Bt+5C-3At^3-3Bt^2-3Ct-3D=-2t^3+4t^2"
    Why -2Bt????? it should be +2Bt
    Once you've done this, enter these expressions into your rleft-hand side; gather together EVERY t^3 term, t^2-term and so on.
    Only then set them equal to the corresponding terms on your right-hand side
     
  4. Oct 5, 2004 #3
    Awesome, thanks alot. Did you get D=116/9?..if you worked it out.
     
  5. Oct 5, 2004 #4

    arildno

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    I didn't.
    Frankly, I'm a bit dubious about how you proceeded further on; could you post your calculations of the coefficients?
    (Or alternatively, check that your "solution" actually IS a solution of your diff.eq.)
     
  6. Oct 5, 2004 #5
    I plugged in the values of A, B, C, D into the equation and I get 0=0.

    For the calculation of B I did: 10t^2-3Bt^2=4t^2 -> B=2
    For the calculation of C I did: 8t+20t-3Ct=0 -> C=28/3
    For the calculation of D I did: -8+140/3-3D=0 -> D=116/9

    The mistake you pointed out earlier about +2Bt only affected the answers for C and D.
     
  7. Oct 5, 2004 #6

    arildno

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    Ok, I didn't read what you wrote carefully enough..:redface:
     
  8. Oct 5, 2004 #7

    HallsofIvy

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    I think there's a fairly obvious mistake here! How did the coefficient of B get to be negative in these last two?
     
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