Form of particular solution for y''-2y'+y=(e^2)/x

[filter54321]

Homework Statement

Find the general solution for:
y''-2y'+y=e^x/x

Homework Equations

NONE - not an initial value problem

The Attempt at a Solution

Solve the homogeneous first:
r²-2r+1=0
r=1 as a double root

So:

y₁=c₁e^x
y₂=c₂xe^x

...but what in God's name is the form for the particular Y (based on the right side of the equation?). I suppose it's the "hard part" of this exercise but I've tried a couple and still don't see it.

 

[LCKurtz]

Do you type that right and really mean the right side is

$$\frac {e^2} x$$

If so, the method would be variation of parameters.

 

[filter54321]

Right side corrected, should be y''-2y'+y=(e^x)/x

 

[LCKurtz]

You need to use variation of parameters to get a particular solution. Here's a link that has a couple of worked examples you can follow if your text isn't sufficient.

http://www.ltcconline.net/greenl/courses/204/ConstantCoeff/variationOParameters.htm