Variation of parameters question

[JamesGoh]

Homework Statement

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

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

Homework 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

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 I am using the method correctly, please feel free to point me in the right direction

 

[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.

 

[R1ckr()11]

What are the solutions to x^2y''- 4xy'+ 6y= 0? that should be your first step.

Yes but which one comes first?

Spoiler

y1=x^3 and y2=X^2 or y1=x^2 and y2=X^3

 

[STEMucator]

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.

 

[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:
u&amp;#039;x^2+ v&amp;#039;x^3= 0 and 2u&amp;#039;x^3+ 3v&amp;#039;x^4= x^4sin(x) that we can solve &quot;algebraically&quot; for u&#039; and v&#039;. <br /> <br /> If we multiply the first equation by 2x and subtract it from the second, the u&#039; terms cancel and we have v&amp;#039;x^4= x^4sin(x). That is the same as v&amp;#039;= sin(x) so that v(x)= -cos(x).<br /> <br /> If we multiply the first equation by 3x and subtract it from the second, the v&#039; terms cancel and we have -u&amp;#039;x^3= x^4sin(x). That is the same as u&amp;#039;= -xsin(x). Integrating by parts, u= xcos(x)- sin(x). <br /> <br /> That is, <br /> y(x)= Cx^3+ Dx^2+ x^3 cos(x)- x^2sin(x)- x^3cos(x)= Cx^3+ Dx^2- x^2sin(x).