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Initial value inhomogeneous differential equation

  1. Oct 20, 2011 #1
    1. The problem statement, all variables and given/known data

    In this problem, we will solve the initial value inhomogeneous differential equation in two steps.
    y''+4y=2x*e^(3x) y(0)=0 y'(0)=0

    1.) First determine yp the particular solution to the differential equation.
    2.) Next determine y=yc+yp the complete solution to the differential equation.


    3. The attempt at a solution
    the roots i got were 0 and -4.
    i did y''+4y=0
    'equivalent' to: r^2 + 4r = 0
    dont know what to do next. usually we'll go onto the yc part but idk what its supposed to look like. the notes suggest that equation will have sin(something) and a cos(something)

    help please! i should be able to do part 2 on my own; but we'll see
     
  2. jcsd
  3. Oct 20, 2011 #2

    Mark44

    Staff: Mentor

    No. The roots of the equation r2 + 4 = 0 are not 0 and -4.
     
  4. Oct 20, 2011 #3
    oh.... i see what i did wrong there. i see it is 4 instead of 4r. the roots are +2i and -2i
     
  5. Oct 20, 2011 #4

    Mark44

    Staff: Mentor

    So the solution to the homogeneous problem will involve sin(2x) and cos(2x).

    Do you have any thoughts on a particular solution?
     
  6. Oct 20, 2011 #5
    what do you mean homogenous? the problem calls is an inhomogenous one?

    but, an example in the book shows y''+4y= e^(3x). when they did yp, the did yp=Ae^(3x)
    but how would mine be since i have that 2x multiplied by my e^(3x) ?
     
  7. Oct 20, 2011 #6

    Mark44

    Staff: Mentor

    The related homogeneous problem is y'' + 4y = 0. sin(2x) and cos(2x) are solutions to that problem.

    If the nonhomogeneous problem happened to be y'' + 4y = 1, you would try a particular solution of yp = A, and then you would figure out what A needed to be.

    If the problem happened to be y'' + 4y = 2x, you would try a particular solution of yp = A + Bx.

    If it were y'' + 4y = 5x2, you would try a particular solution of yp = A + Bx + Cx2.

    The example in your book was y'' + 4y = e3x, so you try yp = Ae3x.

    Now with a right side of 2xe3x, what might you try for a particular solution?
     
  8. Oct 20, 2011 #7
    i dont know =[ .... Ae^(3x)+Bx?
     
  9. Oct 20, 2011 #8

    Mark44

    Staff: Mentor

    How about (A + Bx)e^(3x)?
     
  10. Oct 20, 2011 #9
    Yea ok that makes sense. So what now.. I set A= something and Bx=something to get the value if A and B right?
     
  11. Oct 20, 2011 #10

    lurflurf

    User Avatar
    Homework Helper

    You need to find a and b such that
    y=(A + Bx)e^(3x) is a solution of the equation
    that is
    y''+4y=2x*e^(3x)
    or
    [(A + Bx)e^(3x)]''+4(A + Bx)e^(3x)=2x*e^(3x)
     
  12. Oct 20, 2011 #11
    so it would be:
    3e^(3x) * (2B+3A+3Bx) + 4*(A+Bx)e^(3x) = 2x*e^(3x)
    after some simplifying i ended up getting 6B+13A=0 and 13Bx=2x
    so B= 2/13 and A= -12/169
     
  13. Oct 21, 2011 #12

    Mark44

    Staff: Mentor

    You don't need to ask us - you can check it for yourself. Is (-12/169 + 2/13*x)e^(3x) a solution of y'' - 4y = 2xe(3x)?
     
  14. Oct 21, 2011 #13
    i got the answer actually. thanks for all the help !!!
     
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