Question

how to find y, if y''+2y'+y=x(e^(-x)-cosx) ?

Answer 1

Have any idea? :)

Answer 2

relate it to r^2 +2r +1 = 0 ; for the homogenous part

Answer 3

then relate that back to some structure for the particular part

Answer 4

since the characteristic equation is a perfect square; we get something resulting in:

A e^rx + Bx e^rx

i think

Answer 5

and then ? :)

Answer 6

Solve for the homogeneous case, then find the particular solution, by undetermined coefficients.

Answer 7

hmm.. how to do it..?

Answer 8

just since I need the practice :) our roots are r = -1,-1

\[y_h=c_1e^{-x}+c_2xe^{-x}\]

if i recall it correctly, we can take this and work it for the yp

\[y_p=Ae^{-x}+Bxe^{-x}\]
\[y'_p=A'e^{-x}-Ae^{-x}+B'xe^{-x}-Bxe^{-x}\]
we take the A and B here and equate it to 0 and down the rest
\[y'_p=-Ae^{-x}-Bxe^{-x}\]
\[y''_p=-A'e^{-x}+Ae^{-x}-B'xe^{-x}+Bxe^{-x}\]
\[+2y'_p=-2Ae^{-x}-2Bxe^{-x}\]
\[+y=Ae^{-x}+Bxe^{-x}\]
----------------------------
\[-A'e^{-x}-B'xe^{-x}=xe^{-x}-xcos(x) \]

now we should reintroduce the A'B' stuff the we zeroed out and see what we can come up with

Answer 9

\begin{array}l
A'.e^{-x}+B'.xe^{-x}=0\\
-A'.e^{-x}-B'.xe^{-x}=x(e^{-x}-cos(x))\\
\end{array}

using a cramer rule we can see what happens.

\[B'=\frac{xe^{-x}(e^{-x}-cos(x))}{-xe^{-2x}+xe^{-2x}}\]

ack!!, looks like i ended up with a zero in a denominator.

Answer 10

undetermined coeffs fer sure

Answer 11

this might be more useful than the stuff im trying to remember :)

http://www.wolframalpha.com/input/?i=+y%27%27%2B2y%27%2By%3Dx%28e%5E%28-x%29-cosx%29

Answer 12

ahh, it took and split up the particulate solution into xe^-x and -xcos(x) so we had to work it out twice

Answer 13

good work