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 $Y(s)$. So, the Laplace transform of the given differential equation is
$$s^2Y(s) - sy(0) - y'(0) + 2[sY(s) -
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