# Laplace problem (new) :|

1. Nov 26, 2005

### georgeh

I have the following function with the following I.C. y(0)=1 & y'(0)=1
y''-2y'+2y=cost
I apply laplace.
i get the following
s^2Y(s)-Sy(0)-y'(0)-2[sY(s)-y(0)]+2Y(s)= s/(s^2+1)
-------
i apply the initial condtions
and i obtain
Y(S)[s^2-2s+2]=s-2+s/(s^2+1)
therefore, Y(S)=(s-2)/((s-1)^2+1)+S/[(s-1)^2(s^2+1)]
i am not sure how to procede forward.. any help would be appreciated.
sorry for posting and asking so many questions in such a short period.

2. Nov 27, 2005

### benorin

3. Nov 27, 2005

### benorin

Here you go

From

$$Y(s)=\frac{s-2}{\left( s-1\right)^{2}+1} + \frac{s}{\left( s-1\right)^{2} \left(s^{2}+1\right)}$$

Deal with the terms seperately: first put

$$\frac{s-2}{\left( s-1\right)^{2}+1} = \frac{s-1}{\left( s-1\right)^{2}+1}-\frac{1}{\left( s-1\right)^{2}+1}$$

so that the right-hand side consists of terms on the table;

next, use partial fraction decomposition on the other term, that is write

$$\frac{s}{\left( s-1\right)^{2} \left(s^{2}+1\right)} = \frac{A}{s-1} + \frac{B}{\left( s-1\right)^{2}} + \frac{Cs+D}{s^{2}+1}$$

now cross-multiply to get

$$s = A\left( s-1\right) \left( s^{2}+1\right) + B\left( s^{2}+1\right) + \left( Cs+D\right) \left( s-1\right)^{2}$$

Solve for A,B,C, and D as follows:

If s=1, then $1=2B$;

If s=0, then $0=-A+B+D$;

If s=-1, then $-1=-4A+2B-4C+4D$;

If s=2, then $2=5A+5B+2C+D$;

Solving the above system of equations gives
$$A=0,B=B=\frac{1}{2},C=0,D=-\frac{1}{2},$$

and hence

$$\frac{s}{\left( s-1\right)^{2} \left(s^{2}+1\right)} = \frac{1}{2\left( s-1\right)^{2}} - \frac{1}{2\left( s^{2}+1\right) }$$

All togeather we have

$$Y(s) = \frac{s-1}{\left( s-1\right)^{2}+1}-\frac{1}{\left( s-1\right)^{2}+1}+ \frac{1}{2\left( s-1\right)^{2}} - \frac{1}{2\left( s^{2}+1\right) }\right]$$

hence

$$y(t) = L^{-1}\left\{ Y(s)\right\} = e^{t}\cos(t) - e^{t}\sin(t) + \frac{1}{2}te^{t} - \sin(t)$$

Last edited: Nov 27, 2005
4. Nov 27, 2005

### georgeh

I made a mistake in copying the the laplace transformation

Y(s)= (s-2)/((s-1)^2+1)+ s/(s^2+1)[(s-1)^2+1]

I understood how you the first part of the laplace transform is a form i can look up, but the next part, the (s-1)^2+1, not sure how to deal with that one with PFDC..

5. Nov 28, 2005

### HallsofIvy

Staff Emeritus
Your table certainly has a formula for the inverse Laplace transform of
$\frac{1}{s^2+ 1}$ and also should tell you what to do with "s-1" rather than s.