Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
ranyai12
Group Title
Use Laplace transforms to solve the initial value problem
x''+4x'+8x=2e^(−t) x(0)=0 x'(0)=4
 one year ago
 one year ago
ranyai12 Group Title
Use Laplace transforms to solve the initial value problem x''+4x'+8x=2e^(−t) x(0)=0 x'(0)=4
 one year ago
 one year ago

This Question is Closed

richyw Group TitleBest ResponseYou've already chosen the best response.0
are you allowed to use a table?
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
this is my work and im not sure what i did wrong
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
and yes i am
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
i'm already confused lol. where did you get the 18 from?
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
ooops thats supposed to be an 8
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
yeah if you put an 8 it looks pretty straightforward as you clearly know what to do. I never finished it though as I am cramming for calculus haha!
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
id you get x=(4s+6)/(s=1)(s^2+4a+8)
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
that part always confuses me so if you can please check that id be fine
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
yes I got there that's where I quit. I'm thinking about it again now
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
ok thanks
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
hmm after my partial fractions I'm getting one that is hard to fix into a nice way to inverse laplace
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
what did you end up getting?
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
I can't figure out the answer. I really do apologize but I have a big final coming up and rushed through it. I don't mean to leave you hanging!
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
its ok thanks i just wanted what x= to anyway not the whole thing but I totally understand!!! Finals suck!
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.0
x'' + 4x' +8x= 2*exp(t), x(0)=0, x'(0)= 4 \[ (s^2+4s+8)L[x] (s+4)*0  (1)*4 = \frac{2}{s+1} \] \[ (s^2+4s+8)L[x] = \frac{2}{s+1} +4 \] \[ L[x]= \frac{4s+6}{(s+1)(s^2+4s+8)} \] partial fraction expansion \[ L[x]= \frac{2}{5}\frac{1}{(s+1)} \frac{2}{5}\frac{s7}{s^2+4s+8} \] complete the square on the denominator of the 2nd term to match the laplace tables rewrite the numerator \[ L[x]= \frac{2}{5}\frac{1}{(s+1)} \frac{2}{5}\frac{s+29}{(s+2)^2+2^2} \] rewrite the last term as 2 fractions that match the Laplace tables. \[ L[x]= \frac{2}{5}\frac{1}{(s+1)} \frac{2}{5}\frac{s+2}{(s+2)^2+2^2}+\frac{9}{5}\frac{2}{(s+2)^2+2^2} \] \[ x= \frac{2}{5}e^{t} \frac{2}{5}e^{2t}cos(2t)+\frac{9}{5}e^{2t}sin(2t) \] see http://en.wikipedia.org/wiki/Laplace_transform#Table_of_selected_Laplace_transforms
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.0
I just gave the highlights, as this takes quite a bit of work. Ask if you have any questions.
 one year ago
See more questions >>>
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.