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fixed LT mistake

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Sasserisop 2023-10-20 08:32:37 -06:00
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content/Laplace transform (lec 14-16).md
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@ -49,7 +49,7 @@ $\mathcal{L}\{e^{\alpha t}\sin(bt)\}=\frac{b}{(s-a)^2+b^2}$ these properties are
What if we calculate the LT of $f'$ ? What if we calculate the LT of $f'$ ?
using integration by parts: using integration by parts:
$\mathcal{L}\{f'(t)\}(s)=\int _{0}^\infty e^{-st}f(t)\, dt=e^{-st}f(t)|_{t=0}^{t\to \infty}+s \underbrace{ \int e^{-st}f(t) \, dt }_{ F(s) }$$=sF(s)-f(0)$ $\mathcal{L}\{f'(t)\}(s)=\int _{0}^\infty e^{-st}f'(t)\, dt=e^{-st}f(t)|_{t=0}^{t\to \infty}+ \underbrace{s \int e^{-st}f(t) \, dt }_{ sF(s) }$$=sF(s)-f(0)$
$\mathcal{L}\{f''\}=s^2F(s)-sf(0)-f'(0)$ $\mathcal{L}\{f''\}=s^2F(s)-sf(0)-f'(0)$
in general, we can use proof by induction to show the following (try at home!): in general, we can use proof by induction to show the following (try at home!):
$\mathcal{L}\{f^{(m)}\}=s^mF(s)-s^{m-1}f(0)-\dots-f^{m-1}(0)$ $\mathcal{L}\{f^{(m)}\}=s^mF(s)-s^{m-1}f(0)-\dots-f^{m-1}(0)$