The Laplace transform of $y'(t)$ is $sY(s) - y(0)$ and the Laplace transform of $y(t)$ is $Y(s)$. The Laplace transform of $e^{-4t}$ is $\frac{1}{s+4}$. So, the Laplace transform of the given differential equation is:
$$
sY(s) - y(0) + 4Y(s) = \frac{1}{s+4}
$$
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