Ordinary Differential Equations/Laplace Transform

Definition
Let $$f(t)$$ be a function on $$[0,\infty)$$. The Laplace transform of $$f$$ is defined by the integral


 * $$F(s) = \mathcal{L}\{f\}(s) = \int_0^{\infty} e^{-st} f(t) dt\,.$$

The domain of $$F(s)$$ is all values of $$s$$ such that the integral exists.

Linearity
Let $$f$$ and $$g$$ be functions whose Laplace transforms exist for $$s > \alpha$$ and let $$a$$ and $$b$$ be constants. Then, for $$s > \alpha$$,


 * $$\mathcal{L}\{af + bg\} = a \mathcal{L}\{f\} + b \mathcal{L}\{g\}\,,$$

which can be proved using the properties of improper integrals.

Shifting in s
If the Laplace transform $$\mathcal{L}\{f\}(s) = F(s)$$ exists for $$s > \alpha$$, then


 * $$\mathcal{L}\{e^{at} f(t)\}(s) = F(s - a)\,$$

for $$s > \alpha + a$$.

Proof.

$$ \begin{align} \mathcal{L}\{e^{at} f(t)\}(s) &{} = \int_0^{\infty} e^{-st} e^{at} f(t) dt \\ &{} = \int_0^{\infty} e^{-(s-a)t} f(t) dt \\ &{} = F(s - a)\,. \end{align} $$

Laplace Transform of Higher-Order Derivatives
If $$F(s) = \mathcal{L}\{f(t)\}$$, then $$\mathcal{L}\{f'(t)\} = sF(s) - f(0)$$
 * Proof:
 * $$\mathcal{L}\{f'(t)\} = \int _0^\infty f'(t)e^{-st}dt$$
 * $$= \lim_{C\to\infty}\int _0^C f'(t)e^{-st}dt$$
 * $$= \lim_{C\to\infty} \left.e^{-st}f(t) \right| _0^C - \int _0^C -sf(t)e^{-st}dt$$ (integrating by parts)
 * $$= -f(0) + s \lim_{C\to\infty}\int _0^C f(t)e^{-st}dt$$
 * $$= s\mathcal{L}\{f(t)\} - f(0)$$
 * $$= sF(s) - f(0)\,$$

Using the above and the linearity of Laplace Transforms, it is easy to prove that $$\mathcal{L}\{f''(t)\} = s^2F(s) - sf(0) - f'(0)$$

Derivatives of the Laplace Transform
If $$\mathcal{L}\{f(t)\} = F(s)$$, then $$\mathcal{L}\{tf(t)\} = -F'(s)$$

Laplace Transform of Few Simple Functions

 * 1) $$\mathcal{L}\{1\} = {1 \over s}$$
 * 2) $$\mathcal{L}\{e^{at}\} = {1 \over s-a}$$
 * 3) $$\mathcal{L}\{\cos \omega t\} = {s \over s^2 + \omega^2}$$
 * 4) $$\mathcal{L}\{\sin \omega t\} = {\omega \over s^2 + \omega^2}$$
 * 5) $$\mathcal{L}\{1\} = {1 \over s}$$
 * 6) $$\mathcal{L}\{t^n\} = {n! \over s^{n+1}}$$