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Solutions of Differential Equations with Initial Value Problems

  1. Mar 18, 2006 #1
    Consider the initial-value problem y'''+9y'=0, y(0)=3, y'(0)=-1, y''(0)=2. Given that [​IMG] are three linearly independent solutions of the given homogeneous linear differential equation, determine the particular solution of the initial-value problem.

    So far I have y=C1(1)+C2(cos3x)+C3(sin3x) so 3=C1+C2+C3sin9 Is that right? Please help
  2. jcsd
  3. Mar 18, 2006 #2


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    No, it isn't. You know y(0)=3 and what you want to do is substitute that in to y=C1(1)+C2(cos3x)+C3(sin3x) (which is correct). Doing that gives you
    3 = c1(0) + C2(cos3*0) + C3(sin3*0).
  4. Mar 18, 2006 #3
    OK, so then I would get C2 = 3, Then I would do the derivative of y and get -9sin3x+3c3cos3x, and use y'(0)=-1, so that would be -1=-9sin-3+3c3cos-3
  5. Mar 19, 2006 #4


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    No, what's the derivative of y, forget the C's for a minute.
  6. Mar 19, 2006 #5


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    In y(0)= 3, x= 0, not 3!! That was your mistake in your first post
    sin(0)= 0, cos(0)= 1 so y(0)= 3 gives you C1+ C2= 3.

    In "y'(0)= -1", x= 0, not -1!!!
    3C3= -1.
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