Laplace transformation initial value problem

1. Dec 12, 2008

quietriot1006

1. The problem statement, all variables and given/known data
I need to use the Laplace transformations to solve this IVP

y'-y=2cost(5t) , y(0)=0

2. Relevant equations

3. The attempt at a solution

I seperated the question until the point

Y(s)=(2s)/(s^2+25)(s+1)

and then

2s=A(s+1)+B(s^2+25)

I get the answer for B=-(1/13) and i just want to know how to get A and if i am doing it right so far. ASAP would be great. Thanks.

2. Dec 12, 2008

Staff: Mentor

It would be helpful to see your work in how you got to the Y(s) equation. Also, I assume that the (s + 1) factor is in the denominator, but how you wrote it one could reasonable assume it is in the numerator.
For the 2nd equation above, did you separate the rational expression as A/(s^2 + 25) + B/(s + 1)? If so, that's incorrect. The first expression should be (As + B)/(s^2 + 25), and the other C/(s + 1)

3. Dec 12, 2008

quietriot1006

I messed up the first problem, there is no t after cos. This is how the problem is written.

y'-y=2cos(5t) , y(0)=0

Here is my work from the beginning:
sY(s)-0+Y(s)=(2s)/(s^2+25)

Y(s)=(2s)/((s^2+25)(s+1))

2s=((As+B)(s+1))+(C(s^2+25))

Let s=-1 and then i solve for C which comes out to equal (-1/13). And then I get stuck at A and B. I looked at what you said and you are right and i fixed that but Im still stuck.

4. Dec 12, 2008

rock.freak667

Shouldn't you have sY(s)-Y(s)=2s/(s2+25) ?

5. Dec 12, 2008

quietriot1006

Yes I should. Because of the minus from the initial problem. Thanks for that one. So
C=(1/13) with no minus sign. What about the rest? What number can I choose for s that will cause (s^2+25) to equal zero and then to get A and B?

6. Dec 12, 2008

rock.freak667

try s=0 to get B
then put s=any number to get A

7. Dec 12, 2008

quietriot1006

Cool! i think i got it.
y(t)=(-1/13)cos(5t)+(25/13)sin(5t)+(1/13)e^t

Is this right?

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