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Homework Help: 2nd order nonhomogeneos differential equations with initial conditions

  1. May 16, 2013 #1
    1. The problem statement, all variables and given/known data

    The problem states

    d^2y/dt^2 +15y= cost4t + 2sin t

    initial conditions y(0)=y'(0)=0

    2. Relevant equations

    3. The attempt at a solution

    All I have is this r^2+15=0

    making r(+-)=√15

    and making yh= C1cos√15+C2√15

    the next part includes solve for the nonhomogenous equation

    cos4t + 2sint

    but here is where I can't come with the particular solution can it be

    yp= Acos2t +B sin2t
  2. jcsd
  3. May 16, 2013 #2


    Staff: Mentor

    No, r = ±√(-15) = ±i√(15).

    But rather than working with eit√15 and e-it√15, it's much easier to work with cos(t√(15)) and sin(t√(15)).
    See above.

    What would you have for yh, considering the corrections I made?
    No, these won't work. You'll need four expressions for your particular solution, not two.
  4. May 16, 2013 #3
    So yh= cos(t√(15)) + sin(t√(15))

    and now I need to use the method of undetermined cooficients to come with a particular solution.

    is Yp= Acos 2t +B sin 2t

    Yp'=-2Asin 2t + 2B cos 2t
    Yp''= -4Acos 2t - 4B sin 2t

    right or did I make mistake and if it is right

    it will make

    d^2y/dt^2 +15Yp= -4Acos 2t - 4B sin 2t + 15 Acos 2t +15B sin 2t

    d^2y/dt^2 +15Yp= 11cost + 11sint

    is the particular solution right
  5. May 17, 2013 #4


    User Avatar
    Gold Member

    That is still not the full general solution to the homogeneous equation.

    Could you explain more what you are doing here? In particular why are substituting a form where the argument of sin is 2t?
  6. May 17, 2013 #5


    User Avatar
    Science Advisor

    Your non-homogeneous part has sin(t) and sin(4t). Those do NOT "combine" to give sin(2t).
    Look for a solution of the form A sin(t)+ B cos(t)+ C sin(4t)+ D cos(4t).
  7. May 17, 2013 #6
    so you mean for the nonhomegenous part

    yp=A sin(t)+ B cos(t)+ C sin(4t)+ D cos(4t).
    yp'=A cos(t) - B sin(t) + 4C cos (4t) + - 4D sin (4t)
    yp''= -A sin(t) - B cos(t) - 16C sin(4t) - 16D cos(4t)
  8. May 17, 2013 #7


    Staff: Mentor

    Yes. Now substitute yp and its derivatives into your diff. equation to solve for A, B, C, and D.

    For your general solution you will also need yh, which wasn't right in post #3.
  9. May 17, 2013 #8
    so you mean Yp''-Yp'+15Yp=cost4t to solve

    -A sin(t) - B cos(t) - 16C sin(4t) - 16D cos(4t) + 15A sin(t)+ 15B cos(t)+ 15C sin(4t)+ 15D cos(4t) = cos (4t ) + 2 sin(t)

    -14 A sin(t) +14 B cos(t) -C sin(4t) -D cos(4t) = cos (4t) + 2 sin t

    -C = 1 = -1

    -D = 1 = -1

    -14A = 1 = -1/14

    14B = 1 = 1/14

    are all these values for A B C and D right and my differential equation
  10. May 17, 2013 #9


    Staff: Mentor

    Don't write things like the above. You're saying that -C is equal to 1 and that 1 is equal to -1, which isn't true. You should write something like this
    -C = 1, so C = -1
    -C = 1 ##\Rightarrow## C = -1

    I don't know - check them and see.

    If yp = (-1/14)sin(t) + (1/14)cos(t) - sin(4t) - cos(4t),
    what is yp'' + 15yp? It better be equal to cos(4t) + 2sin(t).
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