Laplace Transform (First Order Differential Equation With Initial Value)

[cas159]

Homework Statement

Y' + 8y = e^-2t*sint, with initial condition y(0) = 0

Homework Equations

L{e^(-2t)sin(t)} = 1/((s+2)²+1)

The Attempt at a Solution

Alright so I've been working on this one for about an hour, I really don't know why but I'm having major problems with these types of problems, whether i don't understand how to set it up or I don't understand partial fraction decomposition, I do not know, but this is what I have done.

(For reference, L{function} is the notation I will use).

L{y''} + 8L{y} = L{e^(-2t)*sin(t)}

sY(s) - y(0) + 8Y(s) = 1/((s+2)² + 1)

1/((s+2)² + 1) = 1/(s²+4s+5)

(s+8)Y(s) = 1/(s²+4s+5)

Y(s) = 1/((s²+4s+5)(s+8))

Now setting up my Partial Fraction Decomposition

As+B/(s²+4s+5) + C/(s+8) = 1/((s²+4s+5)(s+8))

As² + Bs + 8As + 8B + Cs² + 4Cs + 5c = 1

(A+C)s² + (8A + B + 4C)s + 8B + 5C = 1

Setting up my equations:

A + 0B + C = 0
8A + B + 4C = 0
0A + 8B + 5C = 1

I'm getting that A = -1/37, B = 4/37, and C = 1/37, the rest I can do, but the book says that B should be 6/37, can anyone figure out what I did improperly, I would really appreciate it.

 

[Mark44]

Those are the same numbers I get, and I don't see anything you did that was wrong, other than a typo in this line:

L{y''} + 8L{y} = L{e^(-2t)*sin(t)}

That first term should be L{y'}.

Occasionally the answers in the back of the book are wrong, so that might be what happened here. Continue on with your solution, which I get as y(t) = (1/5)e^-8t - (2/5)e^-2tsin(t) - (1/5)e^-2tcos(t). I got this by solving the DE directly, using the method of undetermined coefficients.

 

[cas159]

Sorry I did not get back sooner, thanks a lot for your response, turns out a friend of mine got the same answer and said it was most likely a book error as well. But yes, much appreciated.