Question

@ganeshie8

Answer 1

Help me understand the first part http://puu.sh/lAwD8/4b15d3612a.png \[ty''+y'+ty=0\] in terms of \[(1+s^2)Y'(s)+sY(s)=0\]

Answer 2

\[\frac{ 1 }{ s^2 }[s^2Y(s)-sy(0)-y'(0)]+sY(s)-y(0)+\frac{ 1 }{ s^2 } Y(s)=0\]

Answer 3

thats what i get using laplace

Answer 4

I mean there's no initial conditions either, so how exactly would you approach this

Answer 5

Oh t = 0

Answer 6

laplace transform is not multiplicative

Answer 7

\[\mathcal{L}(fg) \ne \mathcal{L}(f)\mathcal{L}(g)\]

Answer 8

Ah

Answer 9

So we use the s - differential rule!

Answer 10

I think we just need to follow the given instructions

Answer 11

I'm looking at the rules right now

Answer 12

\[ty''+y'+ty=0\]

Answer 13

It says s - differential rules \[\mathcal {L}(tf(t)) = - \frac{ d }{ ds } \mathcal{L} (f(t))\]

Answer 14

which is same as \(-Y'(s)\)

use that

Answer 15

\[-\frac{ d }{ ds }[s^2Y(s)-sy(0)-y'(0)]+sY(s)-y(0)+-\frac{ d }{ ds } Y(s)=0\] you mean this

Answer 16

Yes, that looks good

Answer 17

Ooh I think this does work

Answer 18

yay it works out

Answer 19

Thanks ganeshie

Answer 20

The second part requires integrating factor

Answer 21

Third part not thanks xD

Answer 22

part c is going to be fun

Answer 23

haha ok

Answer 24

So we have to expand the binomial series

Answer 25

yes extended binomial theorem :
\[(1+x)^r = \sum\limits_{n=0}^{\infty} \dbinom{r}{n}x^r\]

Answer 26

where
\[\dbinom{r}{n} = \dfrac{r(r-1)\cdots (r-n+1)}{n!}\]

Answer 27

replace \(x\) by \(s^{-2}\)
\(r\) by \(-1/2\)

Answer 28

**
yes extended binomial theorem :
\[(1+x)^r = \sum\limits_{n=0}^{\infty} \dbinom{r}{n}x^{\color{red}{n}}\]

Answer 29

\[\sum_{n=0}^{\infty}\dbinom{-1/2}{n}s^{-2} = \dfrac{-1/2(-1/2-1)\cdots (-1/2-n+1)}{n!}\] where n = integers

Answer 30

oh oops

Answer 31

forgot about s^(-2)

Answer 32

\[(1+s^{-2})^{-1/2}~~=~~\sum_{n=0}^{\infty}\dbinom{-1/2}{n}s^{-2\color{red}{n}} \]

Answer 33

multiply that by \(s^{-1}\) and try taking the inverse lapalce transform

Answer 34

Does the series become just \[s^{-1}(1+s^{-2})^{-1/2}~~=~~\sum_{n=0}^{\infty}\dbinom{-1/2}{n}s^{-2n-1}\] I'm not sure if I did this right

Answer 35

looks good

Answer 36

Then take inverse laplace of this? Haha

Answer 37

yes

Answer 38

lookup the table for \(\mathcal{L}^{-1}(1/s^{n+1})\)

Answer 39

recall that laplace transofrm is just an integral, so it is linear
so laplace tranform of sum equals the sum of laplace transforms

Answer 40

t^n

Answer 41

\(\mathcal{L}\sum F(s) = \sum \mathcal{L}F(s)\)

Answer 42

there must be some factorial term too

Answer 43

\[\frac{ n! }{ s^{n+1} }\] is t^n

Answer 44

that's the inverse of it

Answer 45

can we say

\(\mathcal{L}^{-1}(1/s^{2n+1}) = \dfrac{t^{2n}}{(2n)!}\)

Answer 46

How did you get that

Answer 47

http://www.wolframalpha.com/input/?i=inverse+laplace+1%2Fs%5E%28n%2B1%29

Answer 48

\(\Gamma(n+1)\) is same as \(n!\)

Answer 49

Oh I don't see that on the table

Answer 50

\[\frac{ n! }{ s^{n+1} }\] is t^n

Answer 51

Correct

Answer 52

stare at it
follows trivially from above

Answer 53

\[\mathcal{L}^{-1}\dfrac{n!}{s^{n+1}} = t^n\]

divide \(n!\) both sides

\[\dfrac{1}{n!}\mathcal{L}^{-1}\dfrac{n!}{s^{n+1}} = \dfrac{t^n}{n!}\]

Answer 54

ohh!

Answer 55

you can pull constants in and out

Answer 56

I didn't think we were allowed to do all that haha

Answer 57

why not, it follows from linearity
\(\int c f(x)\,dx = c\int f(x)\,dx\)

Answer 58

let me put it ina different way so that it may look plausible to you

Answer 59

I meant set them equal and do all that, I see, that makes a lot of sense

Answer 60

\[\mathcal{L}^{-1}\dfrac{n!}{s^{n+1}} = t^n\]


take laplace transform both sides and get
\[ \dfrac{n!}{s^{n+1}} =\mathcal{L} t^n\]

now, divide \(n!\) both sides
\[ \dfrac{1}{s^{n+1}} =\mathcal{L} \dfrac{t^n}{n!}\]

take inverse both sides
\[\mathcal{L}^{-1}\dfrac{1}{s^{n+1}} = \dfrac{t^n}{n!}\]

Answer 61

Yes yes, I get it

Answer 62

This is like the coolest thing ever

Answer 63

you just need to expand the binomial coefficient and mess with factorials a bit

Answer 64

below might help in getting to the required form :
\[(2n)! = 1\cdot 2\cdot 3\cdot 4\cdots (2n-1)\cdot(2n)\\~\\
=[1\cdot 3\cdots (2n-1)][2\cdot 4\cdots (2n)]\\~\\
=[1\cdot 3\cdots (2n-1)][1\cdot 2\cdots (n)]*2^n\\~\\
=[1\cdot 3\cdots (2n-1)]*n!*2^n\\~\\

\]

Answer 65

Ok thank you, I will try