Let $Y(s)$ be the Laplace transform of $y(t)$ and $X(s)$ be the Laplace transform of $x(t)$. Then
$$H(s) = \frac{Y(s)}{X(s)} = \frac{s}{-(s-6)(s+1)}$$
$$Y(s)(s^2 - 5s - 6) = -sX(s)$$
Taking the inverse Laplace transform, we get
$$\frac{d^2y(t)}{dt^2} -
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