Solving Laplace Transforms: x'' + 2x' + 5x = 3e^{-t}cos(2t)

[epheterson]

I hope somebody's up tonight, this is due in the morning and I'm so close.

So I was assigned to solve this differential equation using laplace transforms and although I (think I) can solve it, I'm not getting the same answer that Maple spits out.

The DE is:

$$x'' + 2x' + 5x = 3e^{-t}cos(2t); x(0) = x'(0) = 1$$

Let L(x) = Laplace(x)

So here's my work:
Take the Laplace of everything
$$L(x'')+2L(x')+5L(x) = 3L(e^{-t}cos(2t))$$

Becomes:
$$s^2L(x)-s(1)-(1)+2sL(x)+2(1)+5L(x)=3L(e^{-t}cos(2t))$$

Let L(x) = X

$$X(s^2+2s+5)-s+1=\frac{3(s+1)}{(s+1)^2+4}$$

I solved for X, simplified and broke it into partial fractions to figure out the Inverse Laplase but got the wrong answer. Is there anywhere I messed up in what you can see?

 

[Defennder]

Becomes:
$$s^2L(x)-s(1)-(1)+2sL(x)+2(1)+5L(x)=3L(e^{-t}cos(2t))$$

It should be $$s^2L(x)-s(1)-(1)+2sL(x)-2(1)+5L(x)=3L(e^{-t}cos(2t))$$ instead. You missed the minus sign.

 

[epheterson]

A thank you, took a little while to figure that out

... dang algebra