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Differential Equation- Undetermined Coefficient

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data
    Find a particular solution to the differential equation
    2 y'' - 1 y' - 1 y = -1 t^2 + 2 t + 3 e^{4 t} .


    2. Relevant equations



    3. The attempt at a solution
    I have attempted this problem many times. I think I am having trouble assuming what the general form is.

    assume yp=(At+B)^2+Ce^(4t)=(A^2)(t^2)+2ABt+(B^2)+Ce^(4t)
    yp'=2(A^2)t + 2AB + 4Ce^(4t)
    yp''=2(A^2)+16Ce^(4t)

    So, 2yp''-yp'-yp= 4(A^2)+32Ce^(4t)-2(A^2)t-2AB-4Ce^(4t)-(A^2)(t^2)-2ABt-(B^2)-Ce^(4t)

    So, 27Ce^(4t)=3e^(4t) => C=(1/9)
    -(A^2)(t^2)=-t^2 => A=1
    t(-2(A^2)+2AB) = 2 => B=2
    4(A^2)-2AB-(B^2)=0 .... but it doesn't work.



    I have tried this problem multiple times in multiple different ways this being my last attempt. It's for an online homework and I can't seem to get the right answer, although I feel like I do it right every time...
     
  2. jcsd
  3. Oct 11, 2009 #2

    Mark44

    Staff: Mentor

    Try this instead for your particular solution:
    yp = A + Bt + Ct2 + De4t
    For your general solution, you'll need the particular solution + the solution to the homogeneous problem 2y'' - y' - y = 0. The solution to the homogeneous problem will be yh = c1er1t + c2er2t for some constants r1 and r2 that you need to find.
     
  4. Oct 12, 2009 #3
    I tried that as well:

    yp=A+Bt+Ct^2+De^(4t)
    yp'=B+2Ct+4De^(4t)
    yp''=2C+16De^(4t)

    2yp''-yp'-yp= 4C+32De^(4t)-B-2Ct-4De^(4t)-A-Bt-Ct^2-De^(4t)
    = 27De^(4t)-Ct^2-2Ct-Bt-A-B-4C

    So, 27De^(4t)=3e^(4t)
    D= (1/9)

    -Ct^2=-t^2
    C=1

    -2Ct-Bt=2t
    -2(1)-B=2
    B=-4

    -A-B-4C=0
    -A+4-4(1)=0
    A=0

    SOOoooo.... yp=t^2-4t+(1/9)e^(3t)

    What is wrong with thisss?!?!?! I don't get it. It's getting frustrating.

    Someone shed some light if you could.
     
  5. Oct 12, 2009 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    This should be "+ 4C".

    -A- B+ 4C= 0
    -A+ 4+ 4(1)= 0
    A= 8

     
  6. Oct 12, 2009 #5
    I already tried that answer as well... It's a no go. There must be a glitch in the system or something...
     
  7. Oct 12, 2009 #6

    Mark44

    Staff: Mentor

    Then
    1. Make sure the problem you worked is the same one that was given.
    2. If it is, check your particular solution. It should be true that 2yp'' - yp' - y = -t2 + t + 3e4t. If this equation is an identity, then you have found what the problem asked for. If the system says different, then you have a good case to make that it is wrong. Assuming that the system gives an answer, you should then be able to show that its particular solution is NOT a solution of the given differential equation.
     
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