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2nd order diff eq.

  1. Nov 5, 2009 #1
    1. The problem statement, all variables and given/known data
    find a general solution to the eq:
    y''=sin(3t)+4e^t

    Find a particular solution that satisfies
    y(0) = 1
    y'(1) = 0
    2. Relevant equations



    3. The attempt at a solution
    ive figured the general solution to be
    y(x)= -sin(3t)/9 + 4e^t + c + d
    And thus
    y'(x) = -cos(3t)/3 + 4e^t +c + d

    I know the solution is the general solution + the specific values for c & d.

    so we get 2 equations

    -sin(3(0))/9 +4e^0 +c +d = 1
    4+c+d=1
    c+d=-3

    -cos(3(1))/3 +4e^1 +c +d = 0
    -cos(3)/3 + 4e +c +d =0
    c+d = cos(3)/3 -4e

    now what? if I try substituting values for c & d i end up eliminating both the c and d terms
    eg: c = -3 -d
    (-3 -d ) + d = cos(3)/3 -4e
    -3 = cos(3)/3 -4e ????
     
  2. jcsd
  3. Nov 5, 2009 #2

    Mark44

    Staff: Mentor

    Instead of the way you chose, I would just integrate twice to get y(t).

    You have y''(t) = sin(3t) + 4e^t
    Integrate once to get y'(t) = -(1/3)cos(3t) + 4e^t + C
    Use the fact that y'(1) = 0 to solve for C.
    Now integrate y'(t) to get y(t), remembering to add a different constant of integration, say D. Use the initial condition y(0) = 0 to solve for D.
     
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