IZTECH Math255 Notes p6

By E. Z.

Laplace Transform

Transformation of a differantial equation to a algebraic equation.
\(\ F(s)={∫}^{∞}_{0}f(t)e^{-st}dt \)
\(\ f(t)=e^{at}⇒F(s)=\frac{1}{s-a}=ℒ\{e^{at}\} \)
By comparison test \(\ |f(t)|< ce^{at}, c ∈ℝ⇒∫^∞_0e^{-st}f(t)dt \) converges for \(\ s > a \)
Laplace Transform is homomorphism so: \[\ ℒ\{c_1f(t)+c_2+g(t)\}=c_1ℒ\{f(t)\}+c_2ℒ\{g(t)\} \]

Laplace Transform of Constant Coefficient IVPs

One usefullness of laplace transform is it connects \(\ ℒ\{ y(t) \} \) with \(\ ℒ\{ y'(t) \} \) algebratic equation.
\[\ ℒ\{y'(t) \}=∫^∞_0e^{-st}y'(t)dt=[e^{-st}y(t)]^∞_0-∫^∞_0y(t)(-s)e^{-st}dt \]
If \(\ f(t) \) is continous and \(\ g(t) \) is piecewise continous on interval
\(\ 0 ≤ t ≤ A, A∈ℝ^+ , ∃ K,a,M ∈ ℝ : f(t) ≤ K e^{at}, t ≥ M \text{ for } s > a \)
\(\ ℒ\{f'(t) \}=s ℒ\{f(t)-f(0)\} \)
Generally: \[\ ℒ\{ f^{(n)}(t) \} =s^n ℒ\{ f(t) \}-∑^{n-1}_{k=0}s^{n-1-k}f^{(k)}(t) \]

Inverse Laplace Transform

\[\ ℒ\{f(t)\}=F(s)⇒ℒ^{-1}\{F(s)\}=f(t) \]
\(\ f(t) \& g(t) \) are continous functions \(\ ℒ\{ f(t) \}=F(s) \) & \(\ ℒ\{ g(t) \}=G(s) \)
If \(\ F(s)=G(s) \) for sufficiently large \(\ s \) , then \(\ f(t)=g(t), t > 0 \)
\[\ ℒ^{-1} \{ αF+βG \}=αℒ^{-1} \{ F\}+βℒ^{-1} \{ G \}\]