Question

Need some help please..
Obtain the general solution:
y''-y=e^(-x)[2sinx+4cosx]

Answer 1

This is the problem.. I've been working on these type of problems but I got stuck on this one.
\[y''-y=e^{-x}(2sinx+4cosx)\]

Answer 2

have you made an attempt so far?
it's been a while since i practiced this, but i'll do my best to help.

Answer 3

Yes I know the solution is of the form:
\[y=y _{c}+y _{p}\]

the Yc is:\[Y _{c}=C1e^{-x}+C2e^{x}\]

Answer 4

Im not to sure how to do the Yp..

Answer 5

yep, that looks good (solution of homogeneous eq)
and now we need to do something about the particular solution...

what methods have you generally used for particular solutions? i recall variation of parameters and undetermined coefficients.
the principle of superposition also came to mind.

Answer 6

Yeah thats right, its undetermined coeff..
I think the Yp is:
\[Yp=C3e^{-x}+C4cosx+C5sinx\]
but im not sure..

Answer 7

I know that you take the derivative and then plug in but does this look right?
\[(Ae^{-x}-Bcosx-Csinx)-(Ae^{-x}+Bcosx+Csinx)=e^{-x}(2sinx+4cosx)\]

Answer 8

by the looks of it, i don't think you'll be able to obtain the right-hand side because it is a product of e^x and sin x + cos x on the right, and only e^x, sin x, and cos x individually on the left. the only way a derivative of a function gets you that is if you really start with e^x times sin x + e^x times cos x or the like.

Answer 9

Yeah I see what you mean.. thats why I got confused..

Answer 10

I think my Yp is wrong... i just dont know what it would be..

Answer 11

by product rule, \(e^{-x} \sin x\) has 1st derivative \(-e^{-x} \sin x + e^{-x} \cos x\), notice we never lost that form e^(-x) trig function now. same for 2nd derivative.
i think you could do straight up y_p = Ae^(-x) sin x + Be^(-x) cos x, and see how that works out instead.

Answer 12

Gotcha! Ok let me see... i think its gonna be a big long messy problem haha

Answer 13

yes, and plenty of room to make a small error with taking the derivative of cos x or e^(-x) and getting negative signs allll over. >.>

Answer 14

yep!.. working on it... almost done

Answer 15

Bare with me... i think Ive got it

Answer 16

sure. :)
(checking my own answer as well)

Answer 17

ok so.... moment of truth.... im getting my Yp as:
\[Yp=-2e^{-x}sinx\]
therefore Y:
\[Y=C1e^{-x}+C2e^{-x}-2e^{-x}sinx\]

Answer 18

yep, that is what i got (after i corrected one silly mistake lool)
good job! :)

Answer 19

wheeewww.... awesome! Thanks so much for verifying! Now i can sleep in peace.

Answer 20

haha, no problem!
and gnight!