Question

Here's another second order linear ordinary differential equation that needs to be solved using constant coefficients: y'' + y' -2y = (e^(-2x))((x^2)-1)

Answer 1

Aargh...no:-)

Answer 2

:)

Answer 3

It ain't a good notation I know. :)

Answer 4

http://www.wolframalpha.com/input/?i=solve+y%27%27+%2B+y%27+-2y+%3D+%28e^%28-2x%29%29%28%28x^2%29-1%29

Answer 5

I have the result from wolfram but I need to know how to do it.

Answer 6

what's the answer?

Answer 7

soxhletapparatus Check the wolfram link.

Answer 8

Isn't this one easier than the other one..?

Answer 9

It should be. :S But I can't get theresult...

Answer 10

Where's phi? Was him, wasn't it? I hate computing things...

Answer 11

What did u put for trial?

Answer 12

Anyone who can solve the equation. :)

Answer 13

Did u use 3 terms or 2 for the polynomial?

Answer 14

I can solve for Complimentary but I assume you can as well?

Answer 15

3

Answer 16

and the usual for the exp, so what went wrong?

Answer 17

Yes I can solve the complementary, but the problem is with particular. I also add one X to the (AX^2 + BX + C)e^(-2X) because it is needed so that we don't have same term in complementary and particular solutions. However, I can't get the solution from wolfram.

Answer 18

^^ What I mean is: I multiply the polynomial by X.

Answer 19

OK, u combined everything, I not used to that way, I told u before..

U can try to do separately and premultiply pxe^2x, maybe...

Answer 20

Hmm. Will try to do it separately and report what will happen. Thanks.

Answer 21

Would be good if phi showed up , seems he can do these asleep...:-)

Answer 22

estudeir: It's the same thing whether I do it separately or together. Separately you don't do the second exponential because that term is already contained in the first exponential.

Answer 23

OK, so u r saying the only problem is u can't check your answer in Wolfram, is that it?

Answer 24

No, I can check my answer in wolfram. Problem is it's wrong.I can't get the solution that Wolfram gives.

Answer 25

There has to be some problem with coeffiicients...

Answer 26

That's why I prefer to do separately.....

Answer 27

Actually, I prefer not to do them at all...:-)

Answer 28

Maybe u can post what u have, we can check it instead, I don't mind doing that...

Answer 29

Heh. Where are phi and myininaya ?

Answer 30

Wait so that I can scan the paper...

Answer 31

it is Ax^3+Bx^2+Cx for e^-2x only , not for other e^ax; I wonder why

Answer 32

http://www.mediafire.com/i/?yq564kl3jqd2o9y
Note that the reason there is AX^3 + BX^2 + CX is so that there isn't same term in particular solution as in complementary!

Answer 33

imranmeah91: If you have same terms in particular solution you just merge them.

Answer 34

Don't u need ax^3 + b^x^2 +cx + d....

Answer 35

Dunno, just asking...

Answer 36

Because normally u waould have say, Dexp...

Answer 37

You start out with AX^2 + BX + C times the exponential. Since C times exponential is same as the term in complementary solution you need to multiply whole problem term with x. Hence, AX^3 + BX^2 + CX. I don't think you add D.

Answer 38

OK, multiply exp by x^2 then...

Answer 39

x^2 ? Hmm usually, if there are similar terms in Yc and Yp you multiply with X and again if there are again similar terms. But here multiplying once solves the "similar" problem. Will try it , though.

Answer 40

So long as trial solution is not a solution of homogeneous, procedure should work OK...

Answer 41

Nope. Nothing. First time I got -(1/9) as a coefficient, though. There is -(1/9) in wolfram solution as well. I'm totally in the dark here. If someone can solve this please post it online. As for me, I'm going to eat something. :)