Question

using Laplace: x''+4x'+5x=e^t , x(0)=1 , x'(0)=2


I'm literally just stuck entirely on this problem, and can't find enough resources to work through it.

Answer 1

Lapalace method?!

Answer 2

LaPlace Transform Method

Answer 3

a laplace table would be useful

Answer 4

\[L\{ x''+4x'+5x=e^t\}\]
\[L\{ x''\}+4L\{x'\}+5L\{x\}=L\{e^t\}\]is the easy setup at least :)

Answer 5

I have simplified it down to the inverse Laplace form (I believe) \[(e ^{t} + S + 6)/((S+2)^2-9)\] Then where to?

Answer 6

you cant have a "t" in the outcome there

Answer 7

and if you ike prettier fractions type in frac{a}{b} in teh equation editor; or the latex markups in the reply box

Answer 8

no

Answer 9

Oh, yea. I just noticed that as well. Do I treat the =e^t as the function, and set it equal to zero?

Answer 10

no

Answer 11

1/s-1

Answer 12

treat it as part of the whole shindig

Answer 13

lets go thru the long versions for practice, cause i need the practice lol

Answer 14

Alright.

Answer 15

Firing up LaTeX so I can do Equations

Answer 16

L{x''} = :e^-st x'' = x' e^-st +s :e^-st x'
= x' e^-st +s (e^-st x + s: e^-st x)
= x' e^-st +s (e^-st x + sX(s))
= x' e^-st +se^-st x + s^2X(s) ; (0 , inf)

= s^2X(s) -x'(0) -sx(0)

right?

Answer 17

you may be coming at it from a different angle than me. I could always be wrong though.


I ended up with:
\[L{x}(S ^{2}+4S+5) -S -6 = e^t\]

Answer 18

\[s^2X(s)-s-2+4sX(s)-4+5X(s)=\frac{1}{s-1}\]
\[X(s)(s^2+4s+5)=\frac{1}{s-1}+6+s\]
\[X(s)=\frac{\frac{1}{s-1}+6+s}{s^2+4s+5}\]

Answer 19

that looks 100 times better!

Answer 20

i gotta chk my e^t tho

Answer 21

s+1 not s-1 lol

Answer 22

simplify that mess, and then we can work on undoing it to get a solution

Answer 23

Then I did complete the square to find the bottom.

Answer 24

So right now I have your numerator over (S+2)^2 - 9

Answer 25

\[X(s)=\frac{1+6(s+1)+s(s+1)}{(s+1)(s^2+4s+5)}\]

\[X(s)=\frac{1+6s+6+s^2+s}{(s+1)(s^2+4s+5)}\]

\[X(s)=\frac{s^2+7s+7}{(s+1)(s^2+4s+5)}\]

Answer 26

Yup, that's where I am now. Time for conversion you think?

Answer 27

not yet, we have to split this up into partial fractions in order to better format this into something invertible

Answer 28

I don't think you can.

Answer 29

nvm.

Answer 30

\[\frac{s^2+7s+7}{(s+1)(s^2+4s+5)}=\frac{a}{(s+1)}+\frac{bx+c}{(s^2+4s+5)}\]

\[\frac{-1^2+7.-1+7}{\cancel{(s+1)}(-1^2+4.-1+5)}=\frac{a}{(s+1)}+\frac{bx+c}{(s^2+4s+5)}\]
\[\frac{1-7+7}{1-4+5}=\frac{a=1/2}{(s+1)}+\frac{bx+c}{(s^2+4s+5)}\]

Answer 31

That's right.

Answer 32

Wait, you got A as 1/2? I have it as -2.

Answer 33

double chk your math

Answer 34

im papering it at the moment

Answer 35

i get 4s+18 for the bs+c parts

Answer 36

I hit complex roots -.-

Answer 37

complex roots are fine

Answer 38

I know, but I have had them on every question so far.

Answer 39

hmm, I think i've ended up with:

1/8 e^(-5t)(-e^-4t)+10e^6t-1)

Answer 40

\[\frac18 e ^{-5t}(-e ^{4t}+10e ^{6t}-1)\]

Answer 41

you think thats good?

Answer 42

It looks right, do you have anything?

Answer 43

Plus, I need to eat dinner and the dining hall closes very soon.

Answer 44

yeah,check this out

http://www.stanford.edu/~boyd/ee102/laplace-table.pdf

i think 1/(s-1) was correct to begin with so these might be bad if we used my stuff

Answer 45

the rest was fine tho lol

Answer 46

alright. Well, thanks a million for your help.