Variation of parameters question

1. Sep 3, 2011

JamesGoh

1. The problem statement, all variables and given/known data

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

$x^{2}y" - 4xy' + 6y= x^{4}sin(x)$

2. Relevant equations

$y_{P}=v_{1}(x)y_{1}(x) + v_{2}(x)y_{2}(x)$

$v_{1}(x)'y_{1}(x) + v_{2}'(x)y_{2}(x)=0$

$v_{1}(x)'y_{1}(x)' + v_{2}'(x)y_{2}(x)'=x^{4}sin(x)$

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 $y=Ax^{2}+Bx^{3}-x^{2}sin(x)$

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

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2. Sep 3, 2011

HallsofIvy

You have a lot of general equations with "$y_1$" and "$y_2$" but you don't say what functions they are! You understand, don't you, that $y_1$ and $y_2$ are the two independent solutions to the associated homogeneous equation? What are the solutions to $x^2y''- 4xy'+ 6y= 0$? that should be your first step.

3. Oct 28, 2012

R1ckr()11

Yes but which one comes first?
y1=x^3 and y2=X^2 or y1=x^2 and y2=X^3

4. Oct 28, 2012

Zondrina

First you solve the homogeneous equation :

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

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

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

After this you can apply a nice little theorem involving the wronskian actually to finish the problem off i believe.

5. Oct 30, 2012

HallsofIvy

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 $y= x^r$. $y'= rx^{r-1}$, and $y''= r(r-1)x^{r- 2}$ so that $x^2y''- 4xy'+ 6y= r(r-1)x^r- 4rx^r+ 6x^r= (r^2- 5r+ 6)x^r= 0$. Solving that equation gives r= 3 and r= 2 so that $y_1= x^3$ and $y_2= x^2$ are independent solutions to the homogeneous equation.

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

Putting those into the differential equation, $x^2y''- 4xy'+ 6y= (2u'x^3+ 2x^2u+ 3v'x^4+ 6vx^3)- (8ux^2+ 12vx^4)+ (6ux^2+6vx^3)$$= 2u'x^3+ 3v'x^4= x^4sin(x)$. That means we have the two equations:
$[itex]u'x^2+ v'x^3= 0$ and $2u'x^3+ 3v'x^4= x^4sin(x)$ 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 $v'x^4= x^4sin(x)$. That is the same as $v'= sin(x)$ so that $v(x)= -cos(x)$.

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

That is,
$$y(x)= Cx^3+ Dx^2+ x^3 cos(x)- x^2sin(x)- x^3cos(x)= Cx^3+ Dx^2- x^2sin(x)$$.