Solving initial-value problems with the Laplace transform

For initial-value problems whose initial conditions share an argument of zero, the Laplace transform can be of great utility. The initial conditions are used early on, and you don’t have to insert them into a general solution. Let’s consider

$$y^{\prime\prime}+4y^\prime+4y=0$$

with $y\left(0\right)=1$ and $y^\prime\left(0\right)=0$. We first take the Laplace transform of both sides:

$$\mathscr{L}\left\{y^{\prime\prime}+4y^\prime+4y\right\}=\mathscr{L}\left\{0\right\}.$$

Using

$$\mathscr{L}\left\{f^{\left(n\right)}\left(t\right)\right\}=s^n\mathscr{L}\left\{f\left(t\right)\right\}-s^{n-1}f\left(0\right)-s^{n-2}f^\prime\left(0\right)-\ldots-f^{\left(n-1\right)}\left(0\right),$$

we get

$$s^2\mathscr{L}\left\{y\right\}-sy\left(0\right)-y^\prime\left(0\right)+4\left(s\mathscr{L}\left\{y\right\}-y\left(0\right)\right)+4\mathscr{L}\left\{y\right\}=0.$$

Let’s now use the initial conditions:

$$s^2\mathscr{L}\left\{y\right\}-s+4\left(s\mathscr{L}\left\{y\right\}-1\right)+4\mathscr{L}\left\{y\right\}=0.$$

Solve for $\mathscr{L}\left\{y\right\}$ to get

$$\mathscr{L}\left\{y\right\}=\frac{s+4}{s^2+4s+4}.$$

It remains to take the inverse Laplace transform of both sides. We have

$$\begin{aligned} \mathscr{L}^{-1}\left\{\mathscr{L}\left\{y\right\}\right\}&=\mathscr{L}^{-1}\left\{\frac{s+4}{s^2+4s+4}\right\}\newline y&=e^{-2t}\left(1+2t\right). \end{aligned}$$