Question

I'm having trouble solving a differential equation using Laplace Transforms.

y'''' - 2y'' + y = 0

After applying the Laplace Transforms, I get:

L{y} = 4(s^2 + 1) / (s^2 - 1)^2

But I don't know how to perform the inverse Laplace Transform on it.

Please advise.

Answer 1

u got the formula for the laplace transform for your 2nd and 3rd derivative?

Answer 2

to solve ode with laplace transform u will need initial conditions, y(0), y'(0) and y''(0)

Answer 3

Yeah, y(0) = 0, y'(0) = 4, y''(0) = 0, y'''(0) = 8.

The thing is, I already did those, I just don't know how to do the inverse Laplace. I've looked on wolfram and I saw the answer, but I want to know how to get to it.

Answer 4

u know how to do partial fractions? apply partial fraction and see what u can get?

Answer 5

factor in denominator\[(ax^{2}+bx+c)^{k}=\frac{A _{1}x+B _{1}}{ax^{2}+bx+c}+\frac{A _{2}x+B _{2}}{(ax^{2}+bx+c)^{2}}+...+\frac{A _{k}x+B _{k}}{(ax^{2}+bx+c)^{k}}\]

Answer 6

I'm not sure partial fractions works with a squared term as the denominator, though.

Answer 7

u can factor out your denominator into linear terms b4 applying partial fraction,\[(x^{2}-1)^{2}=(x-1)^{2}(x+1)^{2}\]so now you are solving\[L[y]=\frac{A}{x-1}+\frac{Bs+C}{(x-1)^{2}}+\frac{D}{x+1}+\frac{Es+F}{(x+1)^{2}}\]

Answer 8

i get C = F = 2, other unknowns are 0, so\[L[y]=\frac{2}{(x-1)^{2}}+\frac{2}{(x+1)^{2}}\]

Answer 9

Oh wow, I see. Okay. Thank you so much! :D

Answer 10

Wait, can you explain how you got C = F = 2? I'm getting B + C = 2 and F - E = 2, and also -A + C + D + F = 4.

Answer 11

sry my laptop shutdown for no reason suddenly

Answer 12

wow, a y'''' , someone must really love you

Answer 13

oh wait

Answer 14

im so sorry man, its A B C D only

Answer 15

._.\[\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{x+1}+\frac{D}{(x+1)^{2}}\]so B and D =2, A=C=0.

very sorry for my mistake.

Answer 16

Yup, I got to that step. And @amistre64 at least it's not any higher :P

Answer 17

Im sure that there is a pattern that can be spotted after y'''

y^n = s^n Y(s) - (y(0)+sy'(0)+s^2y''(0)+s^3y'''(0)...s^(n-1)y^(n-1)(0))

but i would have to verify that to be sure

Answer 18

Yeah, that's a theorem. But I've already gone past that step, it's just the partial fraction decomposition that was bothering me.

Answer 19

collect the coefficients for the term with the same power and compare it with 4(s^2 +1)

Answer 20

But with that method, I'm only ending up with 2 solutions. I should have 4, since it's a fourth order differential equation.

I think the answers are e^t, te^t, e^-t, and te^-t.

Also, if the Laplace was \[L\left[ y \right] = 2/(x+1)^2 + 2/(x-1)2\], wouldn't those answers come out to be 2te^t and 2te^-t instead of te^t and te^-t?

Answer 21

for reference purposes here is the solution, though the wolf chose not to use laplace
http://www.wolframalpha.com/input/?i=solve%20y''''-2y''%2By%3D0%2Cy(0)%20%3D%200%2C%20y'(0)%20%3D%204%2C%20y''(0)%20%3D%200%2C%20y'''(0)%20%3D%208&t=crmtb01