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UnkleRhaukus
 3 years ago
\[\mathcal L\big\{f(xa)h(xa)\big\}\]
UnkleRhaukus
 3 years ago
\[\mathcal L\big\{f(xa)h(xa)\big\}\]

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UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0\[\begin{align*} \mathcal L\big\{f(xa)h(xa)\big\}&=\int\limits_0^\infty f(xa)h(xa)e^{px}\text dx\\ \text{let } t=xa\\ \text dt=\text dx\\ x=0\rightarrow t=a\\ x=\infty\rightarrow t=\infty\\ &=\int\limits_{a}^\infty f(t)h(t)e^{p(t+a)}\text dt\\ &= \\ \end{align*}\]

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0\[\begin{align*} &=\int\limits_{a}^\infty f(t)h(t)e^{p(t+a)}\text dt\\ &=e^{pa}\int\limits_{0}^\infty f(t)e^{pt}\text dt\\ &=e^{pa}F(p)\ \\ \end{align*}\] is this right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thats the same thing I show in my book. I just learned this last week. I don't get it either

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The h(xa) or in my book u(ta) as I understand it is just an on off switch for the discontinuous function

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0yeah , h, u , or \(\theta\) are used for the heaviside unit step function

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0\[\mathcal L\big\{f(xa)h(xa)\big\}=e^{pa}F(p)\] \[f(xa)h(xa)=\mathcal L^{1}\big\{e^{pa}F(p)\big\}\]
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