amistre64
  • amistre64
using the Laplace transform; solve: y' + y = sin(3t)
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
i will watch and learn
experimentX
  • experimentX
is y function of t?
amistre64
  • amistre64
yes, dy/dt

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amistre64
  • amistre64
i know the left becoems 3/(s^2+9) the right sides got me a little iffy tho
TuringTest
  • TuringTest
we need the intitial condition y(0) to find the laplace of y'
amistre64
  • amistre64
oh, then y(0)=0 is the condition stated
TuringTest
  • TuringTest
the right side is\[{3\over s^2+9}\]
TuringTest
  • TuringTest
http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table.aspx it's really not too hard
amistre64
  • amistre64
i know it not hard :)
amistre64
  • 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
  • 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
  • TuringTest
brb
experimentX
  • experimentX
answer is given here http://www.intmath.com/laplace-transformation/8-inverse-laplace-solve-de.php
amistre64
  • amistre64
lol, thats the site i pulled it from, yes
anonymous
  • 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
  • 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
  • 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
  • experimentX
i think integration by parts
amistre64
  • amistre64
the ys or the es for the us tho
TuringTest
  • 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
  • amistre64
:e^ny = y e^n/n - :e^n/n y'
amistre64
  • 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
  • amistre64
writers block :)
TuringTest
  • 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
  • TuringTest
I think... I still don't care about the derivation you mentioned though... not yet at least :) ignorance is bliss!
amistre64
  • amistre64
ok, ill find it later, sometime ...
TuringTest
  • TuringTest
oh my answer is wrong anyway, I just realized a huge mistake
amistre64
  • amistre64
and .... yeah, i was going to point tha tout lol
TuringTest
  • 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
  • TuringTest
but again, partial fractions and whatnot
amistre64
  • amistre64
L{y(t)} = Y(s) is the very definitionof the LT right?
TuringTest
  • TuringTest
yeah
amistre64
  • amistre64
is L{y'(t)} = Y'(s) ?
TuringTest
  • TuringTest
no
amistre64
  • amistre64
well, then im halfway there :)
TuringTest
  • 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
  • amistre64
yeah, im wanting to unravel that tablature
TuringTest
  • 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
  • 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
  • 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
  • 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
  • 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
  • TuringTest
yep
TuringTest
  • TuringTest
oh nice proof for the laplace of y' too I get it now :)

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