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Method of undetermined coefficients

  1. Oct 6, 2011 #1
    solve y''+y'+y=cosx-x2ex

    y= yc+yp

    yc:

    characteristic eq gives r2+r+1=0

    using quadratic formula i got r=-1/2+[(sqrt3)/2]i and r=-1/2-[(sqrt3)/2]i

    so yc=e-x/2(c1cos(sqrt3/2)x+c2sin(sqrt3/2)x

    but i have no idea of a particular solution that makes any sense.

    i tried a general approach, the functions and their derivitives on the RHS include terms cosx, sinx, x2ex, xex, ex

    so i good start would be yp=Acosx+Bsinx+Cx2ex+Dxex+Eex

    then find y'p and y''p and sub into original equation.
    but it becomes so dam monstrous there must be another way.
     
  2. jcsd
  3. Oct 6, 2011 #2

    vela

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    You just have to grind through it. You might find it easier to deal with cos x and x2ex separately and then add the two particular solutions together at the end.

    When dealing with the non-trig terms, factor the common exponential out so that you have a polynomial multiplied by the exponential. That way, when you differentiate, you only have to apply the product rule once.
     
  4. Oct 6, 2011 #3
    so i grinded through it and ended up with

    yp=0cosx+1sinx-(1/3)x2ex+(2/3)xex-(4/9)ex

    then to solve i put y=yc+yp

    so y= e-2/x[(c1cos(sqrt3/2)x+c2sin(sqrt3/2)x] + 0cosx+1sinx-(1/3)x2ex+(2/3)xex-(4/9)ex

    yeah?

    that seems mighty ridiculous :(

    and what about c1 and c2?? i dont have initial conditions given so i cant solve for them??
     
  5. Oct 6, 2011 #4

    vela

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    I'm guessing it's just a typo, but the first exponential should be e-x/2, not exp-2/x.

    You're correct. You need initial conditions to find the arbitrary constants.
     
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