Question

y''+9y=cos(2t) where y=y(t) , t>=0

y(0)=1 ; y(-pi/2)=3 ;
solve the differential equation using laplace transforms.

ok so when i use laplace transform on y'' i get s^2Y(s)-sy(0)-y'(0) but i dont have y'(0) in the starting conditions how do i find it ?

Answer 1

Hmm weird D: I've never seen that before...

Answer 2

:( me neither i get that one on exam

Answer 3

@jim_thompson5910 @wio @ranga

Answer 4

I was able to get the solution using the method of undetermined coefficients.
But I still can't figure out how to do using Laplace...
Maybe it has something to do with a time shift or .. something?

Blah it's been too long :( I can't remember this stuff..

Answer 5

i havent covered that method in class , i have to use laplace .. but thanks anyway :)

Answer 6

Try treating it like a constant and seeing what happens.

Answer 7

In other words, like \(c=y'(0)\) and see if you can't solve it in terms of \(c\). Then later on try to solve for \(c\).

Answer 8

\[
(s^2+9)Y-(s+c) = \frac{s}{s^2+4}
\]Which means \[
Y = \frac{s}{(s^2+4)(s^2+9)} +\frac{s+c}{{s^2+9}}
\]

Answer 9

Is it essential at this point that you know whether or not \(c=0,1,2\) etc?

Answer 10

yeah i did that , well i guess not i would split that into s/(s^2+9) and c/(s^2+9) but i would still have C in the solution right ?

Answer 11

Sure. You may be able to use your initial conditions to solve for \(c\).

Answer 12

what i get is :
\[\frac{ 1 }{5}\cos(2t)-\frac{ 6 }{ 5 }\cos(3t)-\frac{ c }{ 3 }\sin(3t)\]

Answer 13

ok thank you i think i can find c now :)

Answer 14

Yeah, that \(-\pi/2\) will make it very easy.

Answer 15

yes thanks a lot ^^

Answer 16

Here is a double check on your inverse Laplace:

http://www.wolframalpha.com/input/?i=inverse+laplace++++s%2F%5B%28s%5E2%2B4%29%28s%5E2%2B9%29%5D+%2B%5Bs%2Bc%5D%2F%5Bs%5E2%2B9%5D

May need to double check my algebra.

Answer 17

Good luck.