using the Laplace transform; solve:
y' + y = sin(3t)

- amistre64

using the Laplace transform; solve:
y' + y = sin(3t)

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

i will watch and learn

- experimentX

is y function of t?

- amistre64

yes, dy/dt

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- amistre64

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

- TuringTest

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

- amistre64

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

- TuringTest

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

- TuringTest

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

- amistre64

i know it not hard :)

- amistre64

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

- TuringTest

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 ;)

- TuringTest

brb

- experimentX

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

- amistre64

lol, thats the site i pulled it from, yes

- anonymous

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...

- amistre64

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

- TuringTest

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) :)

- experimentX

i think integration by parts

- amistre64

the ys or the es for the us tho

- TuringTest

@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...

- amistre64

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

- amistre64

: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

- amistre64

writers block :)

- TuringTest

\[y'+y=\sin(3x)\]\[sY(s)-\cancel{y(0)}^0+Y(s)={3\over s^2+3^2}\]\[Y(s)=\frac1{2s}{3\over s^2+3^2}\]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\[Y(s)=\frac32(\frac16\frac1s-\frac16\frac1{s^2+3})\]\[Y(s)=\frac14\frac1s-\frac1{12}{3\over s^2+3^2}\]\[y(x)=\frac14-\frac1{12}\sin(3x)\]

- TuringTest

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

- amistre64

ok, ill find it later, sometime ...

- TuringTest

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

- amistre64

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

- TuringTest

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

- TuringTest

but again, partial fractions and whatnot

- amistre64

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

- TuringTest

yeah

- amistre64

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

- TuringTest

no

- amistre64

well, then im halfway there :)

- TuringTest

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

- amistre64

yeah, im wanting to unravel that tablature

- TuringTest

\[Y(s)={3\over(s+1)(s^2+9)}={A\over s+1}+{Bs+C\over s^2+9}\]\[A(s^2+9)+(Bs+C)(s+1)=3\]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

- amistre64

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

- TuringTest

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.

- amistre64

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)

- amistre64

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

- TuringTest

yep

- TuringTest

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.