Laplace transform for system of equations

x"(t) + 2x'(t) + x(t) = 0
y"(t) + 2y'(t) + y(t) = x'(t)

X(0)=y(0)=y'(0)=0. X'(0) = 1

I solved for the fist equation and got x = exp(-at)t
This doesn't feel right though because firstly shouldn't there be two solutions to a second order DE and secondly I didn't even use the second equation?

I then solved for the second one using this result and ended up with s/(s+4)^4 for L(y)

Apply L to these equations:
[s^2 X(s) - 0s - 1] + 2 [s X(s) - 0] + X(s) = 0
[s^2 Y(s) - 0s - 0] + 2 [s Y(s) - 0] + Y(s) = s X(s) - 0

Simplifying:
(s^2 + 2s + 1) X(s) = 1
(s^2 + 2s + 1) Y(s) = s X(s).

Since X(s) = 1/(s+1)^2, we obtain (s+1)^2 Y(s) = s * 1/(s+1)^2
==> Y(s) = s/(s+1)^4 = [(s+1)-1]/(s+1)^4 = 1/(s+1)^3 - 1/(s+1)^4.

Inverting yields
x(t) = te^(-t) and y(t) = (1/2!)t^2 e^(-t) - (1/3!)t^3 e^(-t).

I hope this helps!