Question

Find the inverse Laplace transform of
F(s)=ln ((2s^2+8)(s+3)^5)/((s^2-16)(s+2)(s+3)^4)

Answer 1

that is ln( that whole thing)?

Answer 2

if so I bet you will need these laws
ln(ab)=ln(a)+ln(b)
and
ln(a/b)=ln(a)-ln(b)

Answer 3

yes it is

Answer 4

So apply those rules

Answer 5

and then I think the next step would be to differentiate

Answer 6

\[F(s)=\ln\frac{ (2s^2+8)(s+3)^5 }{ (s^2-16)(s+2)(s+3)^4} \]

Answer 7

how do you differentiate

Answer 8

example:
\[F(s)=\ln(\frac{ s+2}{s-5})=\ln(s+2)-\ln(s-5) \\ \frac{dF}{ds}=\frac{1}{s+2}-\frac{1}{s-5} \\ L(tf(t))=-\frac{dF}{ds}=-(\frac{1}{s+2}-\frac{1}{s-5}) \\ t f(t)=-(e^{-2t}-e^{5t})\]

Answer 9

We first need to expand the ln thing using the rules I mentioned above

Answer 10

\[F(s)=\ln(2s^2+8)+5 \ln(s+3)-\ln(s^2-16)-\ln(s+2)-4 \ln(s+3)\]
We could rewrite the part that is the ln(s^2-16) as ln((s-4)(s+4)) as ln(s-4)+ln(s+4)

Answer 11

\[F(s)=\ln(2s^2+8)+5 \ln(s+3)-(\ln(s+4)+\ln(s-4))-\ln(s+2)-4 \ln(s+3) \]
Now differentiate F w.r.t s

Answer 12

\[\frac{ dF }{ ds }=\frac{ 4s }{2s^2+8 }+\frac{ 5 }{ s+3 }-\frac{ 1 }{ s-4 }+\frac{ 1 }{ s-4}-\frac{ 1 }{ s+2 }-\frac{ 4 }{ s+3}\]

does that look correct?

Answer 13

almost

Answer 14

just one sign off

Answer 15

there is a -( ln(s+4)+ln(s-4))

Answer 16

-ln(s+4)-ln(s-4)

Answer 17

\[\frac{ dF }{ ds }=\frac{ 4s }{2s^2+8 }+\frac{ 5 }{ s+3 }-\frac{ 1 }{ s-4 }-\frac{ 1 }{ s-4}-\frac{ 1 }{ s+2 }-\frac{ 4 }{ s+3}\]

Answer 18

now we know L(tf(t))=-dF/ds

Answer 19

\[-\frac{dF}{ds}=-(\frac{4}{2} \frac{s}{s^2+4}+5 \frac{1}{s+3}- \frac{1}{s+4}-\frac{1}{s-4}-\frac{1}{s+2}-4 \frac{1}{s+3})\]

Answer 20

and i missed that there was another sign off

Answer 21

so i corrected it on this newer equation

Answer 22

so you are ready to transform now

Answer 23

so where did u get all that 4/2 and the rest from

Answer 24

I didn't get anything new

Answer 25

\[\frac{4s}{2s^2+8}=\frac{4 \cdot s}{2 \cdot s^2+ 2 \cdot 4} =\frac{4}{2} \frac{s}{s^2+4}\]

Answer 26

I just factored 4 out on top
and factored 2 out on bottom for the first fraction

Answer 27

before we transform, do we need to use partial fractions?

Answer 28

nope not for this there is nothing really to break down

Answer 29

http://www.math.ttu.edu/~gilliam/ttu/s14/chapter_4_lecture_notes.pdf
page 27 is helpful

Answer 30

there is only a couple of rows you need from that table to transform this

Answer 31

\[\frac{ -4 }{ 2 }\cos(2t)+5e^{-3t}+e^{-4t}+e^{4t}+e^{-2t}+4e^{-3t}\]
is that the correct transform?

Answer 32

and you also need to transform the left side

Answer 33

and also 4/2 is just 2

Answer 34

remember we had -dF/ds

Answer 35

according to that chart it can be transformed to t*f(t)

Answer 36

\[t \cdot f(t)=-2 \cos(2t)+5e^{-3t}+e^{-4t}+e^{4t}+e^{-2t}+4e^{-3t}\]

Answer 37

now you can solve for f(t)

Answer 38

\[(-1)^{-2}\cos(2t)\]
does the first term by chance look like anything close to that

Answer 39

why that
(-1)^(-2)
is 1/(-1)^2=1/1=1

Answer 40

first term i see after dividing t on both sides is
\[\frac{-2}{t} \cos(2t)\]

Answer 41

so my professor asked us to use this line from the table to transform\[t^n f(t) \Rightarrow (-1)^nF^{(n)}(s)\]

am not sure how to use that

Answer 42

or \[-2t^{-1}\cos(2t)\]

Answer 43

we actually already used that

Answer 44

\[t^{1}f(t) => (-1)^1 F^{(1)}(s)=-\frac{dF}{ds}\]

Answer 45

that is why we found the derivative of F
and then multiplied it by -1

Answer 46

so we could use that the transformation of that is t*f(t)

Answer 47

\[t^nf(t)=>(-1)^n \frac{d^nF}{ds^n}\]

Answer 48

\[t^1f(t)=tf(t)=> (-1)^1 \frac{d^1F}{ds^1}=(-1)\frac{dF}{ds} \\ t^2f(t)=> (-1)^2 \frac{d^2F}{ds^2}=\frac{d^2F}{ds^2} \\ t^3 f(t)=> (-1)^3\frac{d^3F}{ds^3}=-\frac{d^3F}{ds^3}\]
just trying to show some examples of that thing I started in general

Answer 49

\[F^{(1)}(s) \text{ is the first derivative of } F \text{ w.r.t. } s\]

Answer 50

recall we found that

Answer 51

and multiplied it by -1?

Answer 52

okay that now makes sense because before we did so many steps together that i missed out on what was going on. now i get it. so basically we just divide through by t

Answer 53

yep since we have tf(t) on the left from the -dF/ds part