Question

using the Laplace transform; solve:

y' + y = sin(3t)

Answer 1

i will watch and learn

Answer 2

is y function of t?

Answer 3

yes, dy/dt

Answer 4

i know the left becoems 3/(s^2+9) the right sides got me a little iffy tho

Answer 5

we need the intitial condition y(0) to find the laplace of y'

Answer 6

oh, then y(0)=0 is the condition stated

Answer 7

the right side is\[{3\over s^2+9}\]

Answer 8

http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table.aspx
it's really not too hard

Answer 9

i know it not hard :)

Answer 10

how would we go about Laplacing the left side? and i spose we need to include benign conditions like show all work for points of 80 and all medals given to answers and use your own words to describe how i do this lol

Answer 11

the laplace of y' is\[\mathcal L\{y'(t)\}=sY(s)-y(0)\]which is why I said we need initial conditions
so you have\[sY(s)-y(0)+Y(s)=\frac{3}{s^2+9}\]now solve for Y(s), and then just inverse transform
I'll review how to do that right now so I can look smarter ;)

Answer 12

brb

Answer 13

answer is given here
http://www.intmath.com/laplace-transformation/8-inverse-laplace-solve-de.php

Answer 14

lol, thats the site i pulled it from, yes

Answer 15

seems like a rather long method doesn't it? then again i remember from the video that the guy said that is why ones learns partial fractions...

Answer 16

what is the working behind:\[\int e^{-st}y'(t)dt+\int e^{-st}y(t)dt\to (sY(s)-y(0))+Y\]
im not quite forming a clear picture of that in my head

Answer 17

oh me neither, I don't know the proof of that
I can only do some basic ones with the definition
I usually just work from the tables (don't tell anyone) :)

Answer 18

i think integration by parts

Answer 19

the ys or the es for the us tho

Answer 20

@satellite73 no, it's very short method, aside from the algebra involved\[y'+y=\sin(3x)\]\[sY(s)-\cancel{y(0)}^0+Y(s)={3\over s^2+3^2}\]\[Y(s)=\frac1{2s}{3\over s^2+3^2}\]now we need partial fractions I think...

Answer 21

:e^ny = y e^n/n - :e^n/n y'

Answer 22

:e^n y = e^n Y - :n e^n y might work

:e^n y = e^n Y -n :e^n y

(n+1):e^n y = e^n Y

:e^n y = e^n Y/(n+1) ; [0,inf] = Y/(s-1) ugh

Answer 23

writers block :)

Answer 24

\[y'+y=\sin(3x)\]\[sY(s)-\cancel{y(0)}^0+Y(s)={3\over s^2+3^2}\]\[Y(s)=\frac1{2s}{3\over s^2+3^2}\]now we need partial fractions I think...\[\frac A s+{Bs+C\over s^2+9}\implies A(s^2+9)+(Bs+C)s=\frac32\]blah blah\[s=0:\implies A=\frac16\]\[Cs=0\implies C=0\]\[(A+B)s^2=0\implies B=-\frac16\]so we have\[Y(s)=\frac32(\frac16\frac1s-\frac16\frac1{s^2+3})\]\[Y(s)=\frac14\frac1s-\frac1{12}{3\over s^2+3^2}\]\[y(x)=\frac14-\frac1{12}\sin(3x)\]

Answer 25

I think...
I still don't care about the derivation you mentioned though...
not yet at least :)
ignorance is bliss!

Answer 26

ok, ill find it later, sometime ...

Answer 27

oh my answer is wrong anyway, I just realized a huge mistake

Answer 28

and .... yeah, i was going to point tha tout lol

Answer 29

\[y'+y=\sin(3x)\]\[sY(s)-\cancel{y(0)}^0+Y(s)=(s+1)Y(s)={3\over s^2+3^2}\]of course that was my cat's fault; very distracting animal

Answer 30

but again, partial fractions and whatnot

Answer 31

L{y(t)} = Y(s) is the very definitionof the LT right?

Answer 32

yeah

Answer 33

is L{y'(t)} = Y'(s) ?

Answer 34

no

Answer 35

well, then im halfway there :)

Answer 36

as per formula 35 in the list I linked above\[\mathcal L\{f'(t)\}=sF(s)-f(0)\]which is why I said we needed the initial conditions

Answer 37

yeah, im wanting to unravel that tablature

Answer 38

\[Y(s)={3\over(s+1)(s^2+9)}={A\over s+1}+{Bs+C\over s^2+9}\]\[A(s^2+9)+(Bs+C)(s+1)=3\]blah blah, falling asleep again...
they derive it for the laplace of 1, e^(at), and sin(at) with the definition here
http://tutorial.math.lamar.edu/Classes/DE/LaplaceDefinition.aspx

the last one is kinda tricky, so I would imagine others are quite a feat of integration

Answer 39

they say that the laplace of these things that can be down with other means is more labour intensive; ut there comes a point on these where the laplace and the other methods become just as laborious so the laplace simply makes it simpler to solve

Answer 40

especially when we know the initial conditions in a second-order IVP, (especially if those initial conditions are =0)
that way the whole thing get's reduced to an algebra problem; no calculus required :3
still, the partial fractions and inverse Laplace can get ugly.

Answer 41

i found my concerns from khan acadamy of all places

:e^-st f'(t)dt = e^-st f(t) +s :e^-st f(t) dt ; [0,inf]

e^-inf f(inf) = 0 under the right condition that we will simply assume
-e^0 f(0) = f(0)

s :e^-st f(t) dt = sF(s)

therefore, L{y'} = sY(s) - y(0)

Answer 42

so, this becomes:

sY(s) - 0 +Y(s) = 3\(s^2+9)

Y(s) (s+1) = ....

Y(s) = 3/[(s+1)(s^2+9)]

which is then partialed and inversed

Answer 43

yep

Answer 44

oh nice proof for the laplace of y' too
I get it now :)