# Laplace transform

• Mar 27th 2008, 01:48 AM
Singular
Laplace transform
Using Laplace Transform I was asked to solved this eq.

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

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

thank you(Bow)
• Mar 27th 2008, 01:50 AM
mr fantastic
Quote:

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

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

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

thank you(Bow)

Can you show your work - or at least your answer for Y(s) - to save re-inventing the wheel .....
• Mar 27th 2008, 01:58 AM
Singular
OK wait .... I'll find it first....I forgot where I put it ....(Thinking)
• Mar 27th 2008, 09:01 AM
Singular
Here is my attempt to solve it ...

$
\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 (Headbang) , I don't know what I should do then ...
Anyone can HElp ThankS(Bow)
• Mar 27th 2008, 09:16 AM
TheEmptySet
Quote:

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

$
\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 (Headbang) , I don't know what I should do then ...
Anyone can HElp ThankS(Bow)

The laplace transform of $L(145 \cos(2t))=145 \frac{s}{s^2+4}$

You need to transform the right hand side
• Mar 27th 2008, 09:44 AM
TheEmptySet
I ended up with

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

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

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

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

by partial fractions of the last one we get

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

Combinging like terms we get...

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

Taking the inverse transfrom we get...

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