Ordinary Differential Equations/Laplace Transform

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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.




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[edit]

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.


\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)\,.

Laplace Transform of Higher-Order Derivatives[edit]

If F(s) = \mathcal{L}\{f(t)\}, then \mathcal{L}\{f'(t)\} = sF(s) - f(0)

\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[edit]

If \mathcal{L}\{f(t)\} = F(s), then \mathcal{L}\{tf(t)\} = -F'(s)

Laplace Transform of Few Simple Functions[edit]

  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}}

Inverse Laplace Transform[edit]