Question

Who can help me?

Variation of parameters
\[y''-2y'+y=e^{2x}\]
\[r^2-2r+1=0\]
\[(r-1)^2=0\]
\[r=1\]
\[y_c(x)=c_1e^x+c_2xe^x\]
\[y_p(x)=u_1e^x+u_2xe^x\]
\[y_p'(x)=u_1'e^x+u_2'xe^x+u_1e^x+u_2e^x+u_2xe^x\]
\[u_1'e^x+u_2'xe^x=0\]
\[y_p'=u_1e^x+u_2e^x+u_2xe^x\]
\[y_p''(x)=u_1e^x+u_1e^x+u_2'e^x+u_2e^x+u_2'xe^x+u_2e^x+u_2xe^x\]

Answer 1

@UnkleRhaukus

Answer 2

@radar

Answer 3

@hartnn I'm almost done with the question

Answer 4

glad to see you my friend =D

Answer 5

the next step...everything I did so far seems right

Answer 6

ohhh...I plug them back into the original formula!!!

Answer 7

and then compare the co-efficients.

Answer 8

and then I....? Yeah that's where I'm stuck :P

Answer 9

oh ok

Answer 10

what am I looking for when I compare the coeffients....sorry don't mean to ask dumb questions...but what is my eventual goal? Do I have to manipulate the coeffients if they're diffferent?

Answer 11

sry :S

Answer 12

oh sorry I forgot to write
\[y_p'=u_1e^x+u_2e^x+u_2xe^x\]

Answer 13

Hey, is that the Stewart book?

Answer 14

yes

Answer 15

I solved it using undetermined coefficients and I got \[y=y_c+y_p\]
\[y=c_1e^x+c_2xe^x+e^{2x}\]
and then I was asked to solve the same problem using variation of parameters

Answer 16

you have y_p, y'_p and y''_p

insert them into t your original DE and simplify