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Linear, nonhomogenous, 2nd order ODE IVP

  1. Mar 26, 2007 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    CE: r^2+4
    gs: y=c1 cos(2t) + c2 sin(2t)

    3. The attempt at a solution


    back into problem:

    which becomes:
    5At^2e^t=t^2 then becomes: A=e^-t/5
    4At^et+Bte^t+4Bte^t=0 which becomes: 4A+B+4B=0, B=-(4e^-t)/25

    2Ae^t+2Be^t+Ce^t+4Ce^t=3e^t, e^t's cancel which becomes:
    2A+2B+5C=3 ---> C=[(75e^t)-2)e^-t]/125

    ok, i'm sure i'm not doing this correctly. the book answer is:


    what am i doing wrong? am i skipping a step?
  2. jcsd
  3. Mar 26, 2007 #2
    do you know the annihilator method? Don't waste your time with this guesswork and learn the annihilator method, it will save you from unnecessary stress.
  4. Mar 26, 2007 #3
    When you're working out the particular solution, deal with each term on the right hand side SEPERATELY.
    First consider = t^2, then consider = 3e^t, otherwise you'll confuse yourself.
    i.e. do what you've been doing, except consider them as two seperate cases, and then combine them in the final step.
    Hope that helps.
    Last edited: Mar 26, 2007
  5. Mar 26, 2007 #4


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    Your "guess" is wrong. You dont' have t2e-t on the right hand side, you have t2+ e-t. You should be trying At2+ Bt+ C+ Ee-t.

    Or, as Pseudo Statistic said, handle them separately: try At2+ Bt+ C for t2 and Ee-t for e-t.
  6. Mar 26, 2007 #5
    oh. ok let me try that. is separating them called superposition?

    i don't know the annihilator method, but i'll do some research to find out...would appreciate it if you could explain it if it's right off the top of your head. thanks!
  7. Mar 26, 2007 #6
    ok this is what i did:
    CE: r^2+4
    gs: y=c1 cos(2t) + c2 sin(2t)

    separating: y''+4y=t^2
    yp''=2A becomes 1/4t^2

    solve and get 3/5e^t

    so total is:
    pluggin in the IC i get:


    did i do this correctly? how come i'm not gettin the book answer? i looked through my algebra and it does cancel out to be c1=3/5. am i forgetting to add a constant somewhere?
  8. Mar 27, 2007 #7


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    Staff Emeritus
    Science Advisor

    WHAT becomes 1/4t^2? You want to solve y"+ 4y= t^2. So far you have yp"= 2A and 4yp= 4(At^2+ Bt+ C)= 4At^2+ 4Bt+ 4C so
    yp"+ 4yp= 4At^2+ 4Bt+ (2A+4C)= t^2. For that to be true for all t, you must have 4A= 1, 4B= 0, 2A+ 4C= 0. C is not 0.

    This is correct- you remembered the "4yp"!

    No. Your t^2 part is wrong.

  9. Mar 27, 2007 #8
    oh sorry. i meant after the algebra, A becomes 1/4t^2. i guess maybe my entire mistake is my assumption that C=0. thanks! let me try it again.
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