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Solving DE by undetermined Coefficients

  1. Mar 4, 2014 #1
    1. The problem statement, all variables and given/known data
    y'' + y = xsinx where y(0) = 0 and y'(0) = 1

    2. Relevant equations

    yc = C1sinx + C2cosx

    3. The attempt at a solution
    So my attempt includes the yc which undoubtedly is correct. I now try to solve yp

    yp = Axsinx + Bxcosx
    y''p = A(2cosx - xsinx) + B(-2sinx - xcosx)

    Substituting: y''p + yp = xsinx, clearly doesn't equal xsinx. There are no comparable values to xsinx to even try and solve for A or B.

    Attempting using a different algorithm: yp = (A + Ax)sinx + (B + Bx)cosx
    Notice overlap from yp to yc, therefore multiply by x: yp = (Ax +Ax^2)sinx + (Bx + Bx^2)cosx
    y''p = A(2cosx + 2sinx -xsinx +4xcosx -x^2sinx) + B(-2sinx + 2cosx - 4xsinx - x^2cosx)

    and y''p + yp = A(2cosx + 2sinx + 4xcosx + xsinx) + B(-2sinx + 2cosx - 4xsinx + xcosx)

    Arrgggh this is really troubling me. I honestly have no idea what I'm doing wrong or how to proceed with the Initial Value Problem afterwards.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Mar 4, 2014 #2


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    Try some general linear function times sin(x) & cos(x).

    ##\displaystyle y_p=(Ax+B)\sin(x)+(Cx+D)\cos(x) ##
  4. Mar 4, 2014 #3
    OH my, rookie mistake LOL. but still, i only get that y''p + yp = 2Acosx - 2Csinx. My logic is telling me to try out yp = (Ax^2 + Bx + C)sinx + (Dx^2 + Ex + F)cosx, based upon my classroom notes. I try this and get
    y''p + yp = 2Asinx + 2Dcosx + 2(2Ax+B)cosx -2(2Dx + E)sinx = xsinx

    now i understand that 2(2Ax+B)cosx -2(2Dx + E)sinx = xsinx and so 2(2Ax+B)cosx = 0 and -2(2Dx + E)sinx = xsinx
    as well as: 2Asinx + 2Dcosx = 0, but this seems like an unsolvable puzzle to me. I also believe i may be overthinking things A LOT so um yeah. What's going on here?
    Last edited: Mar 4, 2014
  5. Mar 5, 2014 #4
    Sorry but i edited my post and am not sure if you can see that from outside the thread so.... bump
  6. Mar 5, 2014 #5


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    Your initial guess at the particular solution would normally be ##y_p = (Ax+B)\sin x + (Cx+D)\cos x##; however, because this isn't linearly independent from the homogeneous solution, you need to multiply it by the lowest power of ##x## to make it independent.
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