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Y'' + 5y' = 400 sin(5 t) + 250 cos(5 t), but also has inital conditions? Diff EQ

  1. Feb 25, 2006 #1
    This one got me alittle discombobulated. Initiallly I thought they wanted me to find a particular solution, which I did. But then later I saw it supplied to intial conditions. So I guessed y = Asin(5t)+Bcos(5t) then took derivatives, and applied the intial conditions but also was wrong. Here is the question:

    Find the solution of
    y'' + 5y' = 400sin(5 t) + 250cos(5 t)
    with y(0) = 8 and y'(0) = 8 .
    y =

    Here is my work:

  2. jcsd
  3. Feb 25, 2006 #2


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    Where is the solution of the homogenous part?
  4. Feb 25, 2006 #3


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    y= (8/5)sin(5t)+ 8 cos(5t) is NOT the general solution to the equation. You've left out the solution to the homogeneous part.
  5. Feb 25, 2006 #4
    mr_coffee, remember that it is [itex]y=y_h+y_p[/itex].
  6. Feb 25, 2006 #5
    O yeah i don't know what i was thinking, well I found the homogenous equation added it to the paricular, but still got it wrong:

    For the homogenous i put:
    y = A + Be^(-5t)
    because the other r is 0, e^0 =1.

    8 = y(0) = A+Be^(-5t);
    8 = A+B;

    8=y'(0) = -5Be^(-5t);
    8 = -5B;
    B = -8/5;

    A = 8+8/5 = 48/5;

    So for my answer i put:

    Any ideas wehre i f'ed this one/
  7. Feb 25, 2006 #6
    The inital conditions apply to the total answer, i.e. write down [itex]y=y_h+y_p[/itex] and then apply the IC.
  8. Feb 26, 2006 #7
    I managed to not do it right again, i did what you said correctly i think...
    here is my work:
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