Question

How do I use this linearity, laplace

Answer 1

\[y''-2y'+2y=cost\]

Answer 2

@ganeshie8

Answer 3

\[s^2 \mathcal{L}(y)(s)+2s(y)+2y'(0)-\] yeah I have no clue how to even set this up

Answer 4

Initial conditions are y(0)=1, y'(0)=0

Answer 5

Oh wait that's a formula

Answer 6

\[y'(t)\leadsto sY(s)-y(0)\]
\[y''(t)\leadsto s^2Y(s)-sy(0)-y'(0)\]

Answer 7

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

Answer 8

and y

Answer 9

f and y

two varuables are confusing... maybe stick to one variable

Answer 10

Then put the 2's

Answer 11

Let me rewrite it because that looks weird

Answer 12

\[y''-2y'+2y=\cos t\]

taking laplace tranform both sides gives
\[s^2Y(s)-sy(0)-y'(0)-2(sY(s)-y(0))+2Y(s)=\dfrac{s}{s^2+1}\]

Answer 13

Is this always true \[f'' = s^nY(s)-sy(0)-y'(0)\] and then same for f'

Answer 14

s^2

Answer 15

that is the formula which you must derive upfront

Answer 16

you must have formulas for the laplace transforms of first and second derivatives before starting to solve differential equations using laplace transforms

Answer 17

Oh ok, I was just confused by the y'' and y' I didn't know what to put, but I see it's on the table

Answer 18

These are the formulas :
\[y'(t)\leadsto sY(s)-y(0)\]
\[y''(t)\leadsto s^2Y(s)-sy(0)-y'(0)\]

you can derive them using the definition of laplace transform and IBP

Answer 19

may be lets derive the laplace transform of first derivative

Answer 20

\[f^{n}(t) = s^nF(s)-s^{n-1}f(0)-...-f^{n-1}(0)\]

Answer 21

\[\mathcal{L}(f'(t))=?\]

Answer 22

plug that in the definition of laplace transform (integral) and play with it a bit

Answer 23

Into the integral right

Answer 24

The one we were just working on earlier

Answer 25

yes

Answer 26

Oh I see it would give you the fundamental theorem of calc

Answer 27

watch 10 minutes of this
https://youtu.be/qZHseRxAWZ8?t=1324

Answer 28

watch between 20th minute and 30th minute

Answer 29

sL(f(t))-f(0)

Answer 30

you love this guy

Answer 31

So good, dv = f'(t) in this case

Answer 32

a variable constant XD

Answer 33

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

Answer 34

sign mistake nooo

Answer 35

parth help

Answer 36

\[s^2 \mathcal{L}(y)(s)+2s(y)+2y'(0)\]it is a quadratic expression

Answer 37

get out

Answer 38

lol

Answer 39

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

Answer 40

what do you want to do with this

Answer 41

there we go

Answer 42

now we do partial fractions decomposition

Answer 43

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

Answer 44

#wolframtime

Answer 45

lol he does those terrible jokes when he feels good
and he feels good almost all the time

Answer 46

That was a really good video though

Answer 47

im talking about prof matt..

Answer 48

I konow

Answer 49

hey that was a beautiful gif where did it go

Answer 50

Ok I think I should be good, just needed to know how to set it up, so couldn't we do this with undetermined coeff, variation of parameter

Answer 51

Why exactly is laplace good that all the other methods included what I mentioned and series can't do

Answer 52

all the methods you have included are also good

Answer 53

they do the job pretty well too

Answer 54

engineers like laplace transforms because they always know initial conditions and laplace transforms give the answer one shot

Answer 55

and it's more plug and chug

Answer 56

compare that with the pain of finding complementary and parituclar solutions which are required with all other methods thats you have listed earlier

Answer 57

trueeee

Answer 58

undetermined coeff is probably the worst

Answer 59

Also the method itself is just awesome.
It changes the problem of solving "differential equation" to a problem of solving "algebraic equation"

Answer 60

because you have to guess

Answer 61

haha yeah!

Answer 62

\[y''-2y'+2y=\cos t\]

taking laplace tranform both sides gives
\[s^2Y(s)-sy(0)-y'(0)-2(sY(s)-y(0))+2Y(s)=\dfrac{s}{s^2+1}\]

do take time time to appreciate how easy it is to solve \(Y(s)\) in above equation

Answer 63

ofcourse finding inverse laplace transform is again a pain

Answer 64

Factor and isolate, then use inverse

Answer 65

Yeah

Answer 66

IS this right, \[\frac{ s^3-2s^2+2s-2}{ (s^2+1)(s^2-2s+2) } = \frac{ As+B }{ (s^2+1) }+\frac{ Cs+D }{ (s^2-2s+2) }\] or do I have to go back into my calc text

Answer 67

s^2-2s+2 can be done by completing the square I guess ill have to do that for the final solution but thats easy

Answer 68

just double check numerator once
i don't have pen and paper wid me as of now..

Answer 69

I think it's good, let me see if i get the final answer as they have in the back

Answer 70

im getting s^3-2s^2+s-2 in the numerator

Answer 71

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

Answer 72

ohk my mistake... pls keep going :)

Answer 73

Is there no formula for \[\frac{ 2 }{ s^2+1 }\]

Answer 74

http://www.wolframalpha.com/input/?i=inverse+laplace+1%2F%28s%5E2%2B1%29

Answer 75

I just used wolfram to find the coeff

Answer 76

that looks good

Answer 77

Oh good!

Answer 78

Wolfram, our Lord and Savior

Answer 79

finding inverses of these two must be easy