The Laplace transform of $y''(t)$ is $s^2Y(s) - sy(0) - y'(0)$, the Laplace transform of $y'(t)$ is $sY(s) - y(0)$, and the Laplace transform of $y(t)$ is $Y(s)$. Using the initial conditions $y(a) = 0$ and $y'(a) = 1$, we get:
$$s^2Y(s) - s*0 - 1 + 5[sY(s) - 0] +
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