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Finding a linear differential equation

  1. Jun 12, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-6-12_19-3-31.png

    2. Relevant equations


    3. The attempt at a solution
    So the general solution is the sum of the null solution (Yn) and particular solution (Yp)

    I believe I just need to write:

    y = e2t + 5e8t - 5 + C

    and then find the derivative of both sides

    y' = 2e2t + 40e8t

    is this correct?
     
  2. jcsd
  3. Jun 13, 2016 #2

    Mark44

    Staff: Mentor

    This is almost the general solution. It would be y(t) = Ae2t + 5e8t - 5.
    The first part is the complementary solution (the solution to the homogeneous diff. equation) -- what you're calling the null solution. The second part is the particular solution, a solution to the nonhomogeneous problem.
    No, and it's nowhere near close.
    What you're supposed to do is find the nonhomogeneous differential equation that has y(t) as its solution.Your textbook should have some examples that are similar to this problem. That should be your first resource.
     
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