Solve for z(t) from the simultaneous equation using Laplace transform

Question

Solve for $z(t)$ from the simultaneous equation using Laplace transform

$$ y' + 2y + 6 \int\limits_0^t z \mathrm{d}t = -2 u(t) \\ y' + z' + z = 0$$

subject to $y(0) = -5$ and $z(0) = 6$.

Answer 1

First note that if $u(t)$ is the Heaveside step function, then

$$\int_0^t z(s)\,dt = u(t)\ast z(t)$$

where $\ast$ denotes convolution. So we can rewrite the system as

$$\left\{\begin{aligned} y^{\prime}+2y + 6u(t)\ast z(t) &= -2u(t)\\ y^{\prime} + z^{\prime} + z &= 0\end{aligned}\right.$$

Noting that $y(0) = -5$ and $z(0) = 6$, we take Laplace transforms of both equations to see that

$$\left\{\begin{aligned} (sY(s)+5) + 2Y(s) + 6\frac{Z(s)}{s} &= - \frac{2}{s}\\ (sY(s)+5) + (sZ(s)-6) + Z(s) &= 0\end{aligned}\right.$$

which simplifies to

$$\left\{\begin{aligned}(s^2+2s)Y(s)+6Z(s) &= -5s-2\\ sY(s) + (s+1)Z(s) &= 1\end{aligned}\right.$$

We now solve for $Z(s)$ by means of elimination; multiplying the second equation by $-(s+2)$ and then adding to the first leaves us with

$$(6-(s^2+3s+2))Z(s) = -4-6s\implies Z(s) = \frac{4+6s}{s^2+3s-4}$$

Note that $\dfrac{1}{s^2+3s-4} = \dfrac{1}{(s+4)(s-1)} = \dfrac{1}{5(s-1)} - \dfrac{1}{5(s+4)}$.

Therefore,

$$Z(s) = \frac{4}{5(s-1)} -\frac{4}{5(s+4)} +\frac{6s}{5(s-1)} - \frac{6s}{5(s+4)}$$

Can you take things from here with finding the inverse Laplace Transform?