1. Laplace transform

Using Laplace Transform I was asked to solved this eq.

$\displaystyle \frac{{d^2 y}} {{dx^2 }} + 4\frac{{dy}} {{dx}} + 13y = 145\cos 2t$ $\displaystyle {\text{ }};y(0) = 9$ and $\displaystyle \left. {\frac{{dy}} {{dx}}} \right|_{t = 0} = 19$

I've try it but I got Y(s) that I couldn't find its inverse ... help

thank you

2. Originally Posted by Singular
Using Laplace Transform I was asked to solved this eq.

$\displaystyle \frac{{d^2 y}} {{dx^2 }} + 4\frac{{dy}} {{dx}} + 13y = 145\cos 2t$ $\displaystyle {\text{ }};y(0) = 9$ and $\displaystyle \left. {\frac{{dy}} {{dx}}} \right|_{t = 0} = 19$

I've try it but I got Y(s) that I couldn't find its inverse ... help

thank you
Can you show your work - or at least your answer for Y(s) - to save re-inventing the wheel .....

3. OK wait .... I'll find it first....I forgot where I put it ....

4. Here is my attempt to solve it ...

$\displaystyle \displaylines{ L(y^{''} ) + 4[L(y')] + 13[L(y)] = 145\cos 2t \cr s^2 Y - sy(0) - y'(0) + 4sY - 4y(0) + 13Y = 145\cos 2t \cr s^2 Y - 9s - 19 - 4sY - 36 + 13Y = 145\cos 2t \cr (s^2 - 4s + 13)Y - 9s - 55 = 145\cos 2t \cr Y = \frac{{145\cos 2t + 9s + 55}}{{s^2 - 4s + 13}} \cr}$

After taht I stuck , I don't know what I should do then ...
Anyone can HElp ThankS

5. Originally Posted by Singular
Here is my attempt to solve it ...

$\displaystyle \displaylines{ L(y^{''} ) + 4[L(y')] + 13[L(y)] = 145\cos 2t \cr s^2 Y - sy(0) - y'(0) + 4sY - 4y(0) + 13Y = 145\cos 2t \cr s^2 Y - 9s - 19 - 4sY - 36 + 13Y = 145\cos 2t \cr (s^2 - 4s + 13)Y - 9s - 55 = 145\cos 2t \cr Y = \frac{{145\cos 2t + 9s + 55}}{{s^2 - 4s + 13}} \cr}$

After taht I stuck , I don't know what I should do then ...
Anyone can HElp ThankS
The laplace transform of $\displaystyle L(145 \cos(2t))=145 \frac{s}{s^2+4}$

You need to transform the right hand side

6. I ended up with

$\displaystyle s^2Y-9s-19+4[sY-9]+13Y=145 \frac{s}{s^2+4}$

$\displaystyle [s^2+4s+13]Y=9s+55+145\frac{s}{s^2+4}$

$\displaystyle [(s+2)^2+9]Y=9(s+2)+37+145\frac{s}{s^2+4}$

$\displaystyle Y=9\frac{s+2}{[(s+2)^2+9]} +\frac{37}{[(s+2)^2+9]}+145\frac{s}{(s^2+4)[(s+2)^2+9]}$

by partial fractions of the last one we get

$\displaystyle Y=9\frac{s+2}{[(s+2)^2+9]} +\frac{37}{[(s+2)^2+9]}+145\left[\frac{1}{145}\frac{9s+16}{s^2+4}-\frac{1}{145}\frac{9s+52}{(s+2)^2+9} \right]$

Combinging like terms we get...

$\displaystyle Y=\frac{3}{[(s+2)^2+9]}+9\frac{s}{s^2+4}+8\frac{2}{s^2+4}$

Taking the inverse transfrom we get...

$\displaystyle y(t)=e^{-2t}\sin(3t)+9\cos(2t)+8\sin(2t)$