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

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richyw
 3 years ago
Best ResponseYou've already chosen the best response.0are you allowed to use a table?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this is my work and im not sure what i did wrong

richyw
 3 years ago
Best ResponseYou've already chosen the best response.0i'm already confused lol. where did you get the 18 from?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ooops thats supposed to be an 8

richyw
 3 years ago
Best ResponseYou've already chosen the best response.0yeah 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!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0id you get x=(4s+6)/(s=1)(s^2+4a+8)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that part always confuses me so if you can please check that id be fine

richyw
 3 years ago
Best ResponseYou've already chosen the best response.0yes I got there that's where I quit. I'm thinking about it again now

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what did you end up getting?

richyw
 3 years ago
Best ResponseYou've already chosen the best response.0I 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!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0its ok thanks i just wanted what x= to anyway not the whole thing but I totally understand!!! Finals suck!

phi
 3 years ago
Best ResponseYou've already chosen the best response.0x'' + 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

phi
 3 years ago
Best ResponseYou've already chosen the best response.0I just gave the highlights, as this takes quite a bit of work. Ask if you have any questions.
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