Question

can someone help with the laplace transform of a differential equation?

Answer 1

\[y''+y=\sin(t)........y(0)=y'(0)=0\]

Answer 2

what does...... between sin(t) and y(0) =y'(0) =0 mean?

Answer 3

i was simply separating the initial conditions because spaces wouldn't work

Answer 4

Use
\;
\quad
\qquad

One of those for spacers :) There are other ways to make spaces, but I forget..

Answer 5

thanks

Answer 6

So we start by taking the Laplace of each side.
We should be able to write \(\Large \mathscr{L}(y'')\) in terms of \(\Large \mathscr{L}(y)\).
Do you remember the identity thing?

Answer 7

yes its \[L(y'')=s^2L(y)-sy(0)-y'(0)\]

Answer 8

Ok cool.
Let's call it something else to abbreviate, like \(\Large \mathscr{L}(y)=Y\)

So we have \[\Large y''+y\quad =\quad \sin t,\qquad \qquad y_o=y_o'=0\]Taking the Laplace gives us:\[\Large s^2Y-s y_o-y_o'+Y\quad=\quad \mathscr{L}(\sin t)\]

Answer 9

So we just need to fill in some stuff and simplify before we can take the inverse Laplace, yes? :o

Answer 10

yeah, so i do all the plugging in and what not and I end up with \[L(y)=1/(s^2+1)^2\] and then I get stuck

Answer 11

hmm

Answer 12

I mean we could maybe do Partial Fraction Decomp?
I don't think that's the route we want to take though :o hmm

Isn't there something for dealing with powers like this?
Something to do with derivative of ummm... grrr I can't remember D:

Answer 13

Grr thinking :d

Answer 14

i think you can do this using convolution

Answer 15

although partial fractions might be easier

Answer 16

i'll look into that

Answer 17

\[\Large \frac{1}{(s^2+1)^2}\quad=\quad \frac{As+B}{s^2+1}+\frac{Cs+D}{(s^2+1)^2}\]

Answer 18

Really? It just looks like such a mess lol
But yah maybe that's the way to go :o

Answer 19

using convolution, the answer can be written as an integral: integrate(sin(z)sin(t-z),{z,0,t})

Answer 20

Hmm yah it seems that Partial fractions doesn't work :(
You end up with A=0, B=0, C=0 and D=1.
Bringing us back to where we started lol.

Answer 21

to evaluate the integral, use the sin addition formula

Answer 22

oh I see

Answer 23

for partial fractions, let f(t)=InverseLaplace{1/(s^2+1)^2},
then f(t/i)=i*InverseLaplace{1/(s^2-1)^2}}

Answer 24

the problem definitely has a solution of (sin(t)-t cos(t))/2

Answer 25

have fun.

Answer 26

:)