A community for students.
Here's the question you clicked on:
 0 viewing
UnkleRhaukus
 3 years ago
\[\mathcal L\big\{f(xa)h(xa)\big\}\]
UnkleRhaukus
 3 years ago
\[\mathcal L\big\{f(xa)h(xa)\big\}\]

This Question is Closed

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?

ChmE
 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

ChmE
 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\[\mathcal L\big\{f(xa)h(xa)\big\}=e^{pa}F(p)\] \[f(xa)h(xa)=\mathcal L^{1}\big\{e^{pa}F(p)\big\}\]
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.