## ranyai12 Group Title Use Laplace transforms to solve the initial value problem x''+4x'+8x=2e^(−t) x(0)=0 x'(0)=4 2 years ago 2 years ago

1. ranyai12 Group Title

@richyw

2. richyw Group Title

are you allowed to use a table?

3. ranyai12 Group Title

this is my work and im not sure what i did wrong

4. ranyai12 Group Title

and yes i am

5. richyw Group Title

i'm already confused lol. where did you get the 18 from?

6. ranyai12 Group Title

ooops thats supposed to be an 8

7. richyw Group Title

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!

8. ranyai12 Group Title

id you get x=(4s+6)/(s=1)(s^2+4a+8)

9. ranyai12 Group Title

that part always confuses me so if you can please check that id be fine

10. ranyai12 Group Title

(s-1)*

11. richyw Group Title

yes I got there that's where I quit. I'm thinking about it again now

12. ranyai12 Group Title

ok thanks

13. richyw Group Title

hmm after my partial fractions I'm getting one that is hard to fix into a nice way to inverse laplace

14. ranyai12 Group Title

what did you end up getting?

15. richyw Group Title

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!

16. ranyai12 Group Title

its ok thanks i just wanted what x= to anyway not the whole thing but I totally understand!!! Finals suck!

17. ranyai12 Group Title

@jagan

18. phi Group Title

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 $L[x]= \frac{2}{5}\frac{1}{(s+1)} -\frac{2}{5}\frac{s-7}{s^2+4s+8}$ complete the square on the denominator of the 2nd term to match the laplace tables rewrite the numerator $L[x]= \frac{2}{5}\frac{1}{(s+1)} -\frac{2}{5}\frac{s+2-9}{(s+2)^2+2^2}$ re-write 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

19. phi Group Title

I just gave the highlights, as this takes quite a bit of work. Ask if you have any questions.