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Variation of parameters question

  1. Sep 3, 2011 #1
    1. The problem statement, all variables and given/known data

    Using the variation of parameters method, find the general solution of

    [itex]x^{2}y" - 4xy' + 6y= x^{4}sin(x)[/itex]

    2. Relevant equations

    [itex]y_{P}=v_{1}(x)y_{1}(x) + v_{2}(x)y_{2}(x)[/itex]

    [itex]v_{1}(x)'y_{1}(x) + v_{2}'(x)y_{2}(x)=0[/itex]

    [itex]v_{1}(x)'y_{1}(x)' + v_{2}'(x)y_{2}(x)'=x^{4}sin(x)[/itex]

    yp is the particular solution, v1,v2, y1 and y2 are nominal functions of x


    3. The attempt at a solution

    see pdfs below

    The tutor's answer is [itex]y=Ax^{2}+Bx^{3}-x^{2}sin(x)[/itex]

    Im not sure if Im using the method correctly, please feel free to point me in the right direction
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Sep 3, 2011 #2

    HallsofIvy

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    You have a lot of general equations with "[itex]y_1[/itex]" and "[itex]y_2[/itex]" but you don't say what functions they are! You understand, don't you, that [itex]y_1[/itex] and [itex]y_2[/itex] are the two independent solutions to the associated homogeneous equation? What are the solutions to [itex]x^2y''- 4xy'+ 6y= 0[/itex]? that should be your first step.
     
  4. Oct 28, 2012 #3
    Yes but which one comes first?
    y1=x^3 and y2=X^2 or y1=x^2 and y2=X^3
     
  5. Oct 28, 2012 #4

    Zondrina

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    First you solve the homogeneous equation :

    [itex]L[y] = x^2y'' - 4xy' + 6y = 0[/itex]

    You should get a pair of solutions. These solutions will allow you to solve the non homogeneous system :

    [itex]L[y] = x^2y'' - 4xy' + 6y = x^4sin(x)[/itex]

    After this you can apply a nice little theorem involving the wronskian actually to finish the problem off i believe.
     
  6. Oct 30, 2012 #5

    HallsofIvy

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    Since this is over a month old, and R1ckr()11 sent me a pm about it:

    You can solve the corresponding homogeneous equation by trying a solution of the form [itex]y= x^r[/itex]. [itex]y'= rx^{r-1}[/itex], and [itex]y''= r(r-1)x^{r- 2}[/itex] so that [itex]x^2y''- 4xy'+ 6y= r(r-1)x^r- 4rx^r+ 6x^r= (r^2- 5r+ 6)x^r= 0[/itex]. Solving that equation gives r= 3 and r= 2 so that [itex]y_1= x^3[/itex] and [itex]y_2= x^2[/itex] are independent solutions to the homogeneous equation.

    Now look for a solution to the entire equation of the form [itex]y= u(x)x^2+ v(x)x^3[/itex]. Differentiating, [itex]y'= u'x^2+ 2ux+ v'x^3+ 3vx^2[/itex]. There are, in fact, an infinite number of solutions of that form so we "narrow the search" by requiring that [itex]u'x^2+ v'x^3= 0[/itex]. That leaves [itex]y'= 2ux+ 3vx^2[/itex].
    Differentiating again, [itex]y''= 2u'x+ 2u+ 3v'x^2+ 6vx[/itex].

    Putting those into the differential equation, [itex]x^2y''- 4xy'+ 6y= (2u'x^3+ 2x^2u+ 3v'x^4+ 6vx^3)- (8ux^2+ 12vx^4)+ (6ux^2+6vx^3)[/itex][itex]= 2u'x^3+ 3v'x^4= x^4sin(x)[/itex]. That means we have the two equations:
    [itex][itex]u'x^2+ v'x^3= 0[/itex] and [itex]2u'x^3+ 3v'x^4= x^4sin(x)[/itex] that we can solve "algebraically" for u' and v'.

    If we multiply the first equation by 2x and subtract it from the second, the u' terms cancel and we have [itex]v'x^4= x^4sin(x)[/itex]. That is the same as [itex]v'= sin(x)[/itex] so that [itex]v(x)= -cos(x)[/itex].

    If we multiply the first equation by 3x and subtract it from the second, the v' terms cancel and we have [itex]-u'x^3= x^4sin(x)[/itex]. That is the same as [itex]u'= -xsin(x)[/itex]. Integrating by parts, [itex]u= xcos(x)- sin(x)[/itex].

    That is,
    [tex]y(x)= Cx^3+ Dx^2+ x^3 cos(x)- x^2sin(x)- x^3cos(x)= Cx^3+ Dx^2- x^2sin(x)[/tex].
     
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