# Thread: Variation of parameters Diff EQ

1. ## Variation of parameters Diff EQ

y" + 2y' + y = 4(e^-x)*ln(x)

Here's my work:

1. r^2 + 2r + 1 = 0
2. (r + 1)^2 = 0; r= -1, -1
3. yc = c1e^(-x) + c2xe^(-x)
4. yp = f1e^(-x) + f2xe^(-x)

So:

f1'e^(-x) + f2'xe^(-x) = 0
-f1'e^(-x) + f2'(e^(-x) - xe^(-x)) = 4(e^-x)*ln(x)

This system simplifies to:

f1' + f2'x = 0
-f1' + f2'(1 - x) = 4*ln(x)

Then I used Cramer's Rule to solve for f1' and f2'. I end up with:

f1' = 4x*ln(x)
f2' = 4*ln(x)

Then I integrate these to find f1 and f2 and plug back into yp = f1e^(-x) + f2xe^(-x). Then the answer is y = yc + yp. However, I don't get the answer in the back of the book. My yc is correct; so my error has to lie in my calculation of the yp. Are my f1' and f2' correct? Any suggestions?

Thanks a bunch,

Jim

2. I used maple to check your work.
I agree you are very close but have lost a minus sign.

I hope you can follow the maple screen shot below.

$\displaystyle f_1'=-4x\ln(x)$

I hope this helps.
Good luck.

3. Thanks Empty Set! Are you a math teacher?

JN