Question

Solve integrodifferential equation

Answer 1

\[f(t)=te^{t}+\int_0^t \tau f(t-\tau)d\tau\]

Answer 2

@jim_thompson5910

Answer 3

i got to this but idk if there is just an error i'm doing in partial fractions

\[\frac{s^2}{(s-1)^2(s^2-1)}\]

Answer 4

kainui you know DE?

Answer 5

I just started learning, I know very little about laplace transforms. I'm afraid I can't really offer any advice here.

Answer 6

you have any clue whats wrong?

Answer 7

actually this might be the problem with the wrng answer

Answer 8

we did a problem that we couldn't find the answer to in class and found out that s^2+1)

Answer 9

wait

\[\frac{s^2}{(s-1)^2(s^2-1)}=\frac{s^2}{(s-1)^3(s+1)}\]

Answer 10

\[\frac{s^2}{(s-1)^3(s+1)}=\frac{A}{s-1}+\frac{B}{(s-1)^2}+\frac{C}{(s-1)^3}+\frac{D}{s+1}\]

Answer 11

\[s^2=A(s-1)^2(s+1)+B(s+1)(s-1)+C(s+1)+D(s-1)^3\]

Answer 12

@mukushla is this the only way to solve this partial fraction

Answer 13

seems like a lot of expansion -.-

Answer 14

yes i think but u can compute \(D\) and \(C\) before doing that messy multiplication

Answer 15

for example for \(D\) multiply both sides by \(s+1\) and let \(s=-1\)

Answer 16

is that allowed?

Answer 17

yeah..

Answer 18

if you ultiply by s+1 on both sides you get

\[\frac{s^2}{(s-1)^3}=\frac{C(s+1)}{(s-1)^3}+D\]

Answer 19

\[D=\frac{1}{-8}\]

Answer 20

For C

\[\frac{s^2}{s+1}=A(s-1)^2+B(s-1)+C+D(s-1)^3\]

Answer 21

C=1/2

Answer 22

u can find A,B now easily by just putting vakues of s,say s=0,s=2....then soloving simultaneously for A,B....C,D values are correct...

Answer 23

let s=1

\[b=1-\frac{3}{8}=\frac{5}{8}\]

Answer 24

\[-A+B=\frac{-3}{8}\]

Answer 25

\[A=1\]

Answer 26

\[\frac{1}{s-1}+\frac{5}{8}\frac{1}{(s-1)^2}+\frac{1}{2}\frac{1}{(s-1)^3}-\frac{1}{8}\frac{1}{s+1}\]

Answer 27

this is not the answer they got

Answer 28

the last two terms are correct

Answer 29

i think u made a little mistake up there...

see this also
http://openstudy.com/users/eliassaab#/updates/500c4d0fe4b0549a89302b3c

Answer 30

her tutorial is vague for a and b

Answer 31

@mukushla

Answer 32

\[\frac{1}{(i-1)^2(i+1)}=\frac{1}{i^2-2i+1(i+1)}\]

Answer 33

that was for showing cover-up method ... dont let x=i

Answer 34

i know cover up method i've never used it as such though i always thought using it the way you told me to manipulated the answer somehow

Answer 35

anyways

\[\frac{s^2}{(s-1)^3(s+1)}=\frac{A}{s-1}+\frac{B}{(s-1)^2}+\frac{\frac{1}{2}}{(s-1)^3}+\frac{\frac{-1}{8}}{(s+1)}\]

Answer 36

let \(s=0\) gives \(-A+B-C+D=0\) or \(B-A=\frac{5}{8}\)
let \(s=2\) gives \(A+B+C+\frac{D}{3}=\frac{4}{3}\) or \(A+B=\frac{7}{8}\)

it gives \(A=1/8\) and \(B=3/4\)

Answer 37

check with this
http://www.wolframalpha.com/input/?i=partial+fractions++s%5E2%2F%28s%5E2-1%29%2F%28s-1%29%5E2

Answer 38

yes that's expansion... and it's insane

Answer 39

it's something i don't want to do if it's possible

Answer 40

how'd you pick 0 and 2 ?

Answer 41

did you just pick two to creat system of equations?

Answer 42

that's one hell of a fraction !!!

Answer 43

yes i've never seen a partial fraction in which expansion is super long and a cover up method as such. Most books say that cover up isn't allowed with repeated factors

Answer 44

o well onto the next chapter -.-

Answer 45

lol ... man i wouldn't even try that without cover up method.

Answer 46

as long as i have this down in two weeks i'll be fine i have like a 95 % and he drops the lowest test so

Answer 47

honestly use wolf!!

Answer 48

it's in the book .... lol it's like the 2nd odd problem too.. the first one was like f(t)=sin(t)... and then this....

Answer 49

differentiating i got
http://www.wolframalpha.com/input/?i=x+y%27%27+-+y%27+%2B+e^x%281-t-t^2%29+%3D+0

Answer 50

i guess something went wrong!!

Answer 51

i guess not the right way Q is asking for.

Answer 52

nope that's correct... a=1/8 b=3/4 c=1/2 d=1/8