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Method of undetermined coefficients Problems

  1. Nov 16, 2009 #1
    been working on these all week some help please..

    9) 3y''-6y'+30y= 15sinx+exp(x)tan(3x)

    10) y''-2y'+y= 4x^2 -3+ Exp(x)/x


    2. Relevant equations

    Variations of parame
    Method of undetermined coefficients

    thoughts:

    i think that you need to split the eqtuaion into 2 parts. Using variation of parameters on 3y''-6y'+30y=exp(x)tan(3x).. and using method of undetermined coefficients on 3y''-6y'+30y=15sinx.. i have tried this approach on both problems with no luck.
     
    Last edited: Nov 16, 2009
  2. jcsd
  3. Nov 16, 2009 #2

    HallsofIvy

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    It's impossible to tell what you should do without knowing what you did and seeing where you ran into trouble. What did you do?
     
  4. Nov 17, 2009 #3
    sorry here is the work i did for #9

    Generic solution:
    y = e^Ax
    y' = Ae^Ax
    y'' = A^2 e^Ax

    3A^2 - 6A + 30 = 0
    3(A^2 - 2A + 10) = 0
    A = [2 +/- sqrt(4 - 40) ]/2 = 1 +/- sqrt(-9) = 1 +/- 3i

    y = C1 e^(1 +/- 3i)x + C2 e^(1 +/- 3i)x
    y = C1 e^(1 +/- 3i)x + C2 e^(1 +/- 3i)x
    y = C1 e^x (cos 3x + i sin 3x) + C2 e^x (cos 3x - i sin 3x)
    y = C1' e^x cos 3x + C2' i e^x sin 3x

    specific soluition using undetermined coefficients.

    15sinx
    If we plug in y = A sin x, we get:

    3y''-6y'+30y= 27 sin x - 6 cos x

    And if we plug in y = B cos x, we get:

    3y''-6y'+30y= 27 cos x + 6 sin x

    With a little vector addition:

    27A + 6B = 15; 9A + 2B = 5
    -6A + 27B = 0; A = 9/2 B

    81/2 B + 2B = 5
    85/2 B = 5
    B = 10/85 = 2/17
    A = 9/17

    So y = 9/17 sin x + 2/17 cos x is the specific solution that will generate 15 sin x.


    I can't quite get the third piece
     
  5. Nov 18, 2009 #4
    This is a damn hairy diffeq. For 9, use variation of parameters. First calculate the homogeneous equation. 3y''-6y'+30y=0. since this has constant coeff, assume that the answer is of the form y=e^(mx) and plugging it in u get a auxiliary equation of the form

    3m^2-6m+30=0. Solving it u get the homogenous eqn y= c[1]*Exp[x]Sin[3x]+c[2]*Exp[x]Cos[3x].

    now that we have the homogeneous eqn, we can use variation of parameters to calculate v1.
    where v1 = Integral[(-y2R(x))/W] for the Wronskian we get 3Exp[2x]. so

    v1 = -Integral[{(Exp[x]Cos[3x])*(15Sin[x]+Exp[x]Tan[3x])}/3Exp[3x]]
    solving that puppy out to take u a while but after Integrating you should get

    v1 = 1/816 E^(4 x) (204 Cos[2 x] - 24 E^x Cos[3 x] - 255 Cos[4 x] -
    408 Sin[2 x] + 40 E^x Sin[3 x] + 255 Sin[4 x])

    then multiply v1*y1 to get the first part of the particular soln.

    then find v2, multiply by y2 etc...

    b/w i used mathematica to calculate v1 and i suggest u do the same for the others unless ur teacher specified otherwise. gl.

    also same process as above for 10.
     
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