anonymous
  • anonymous
(again)Solve by Laplace transform: \[\frac{d^2y}{dx^2} + a \frac{dy}{dx} - 2a^2y=0\] y(0)=6, y'(0) = 0
Differential Equations
katieb
  • katieb
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anonymous
  • anonymous
\[\frac{d^2y}{dx^2} + a \frac{dy}{dx} - 2a^2y=0\]\[sY'-0 +a(sY-6)-2a^2Y=0\]\[s(sY-6) +a(sY-6)-2a^2Y=0\]\[(s^2+as-2a^2)Y=6a+6s\]\[Y=\frac{6a+6s}{(s+2a)(s-a)}\]So far so good? Or...??
experimentX
  • experimentX
|dw:1355844666804:dw|
anonymous
  • anonymous
y(0)=6, y'(0) = 0

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experimentX
  • experimentX
sorry ... i didn't see, looks like the transform of d^2y/dx^2 is incorrect.
experimentX
  • experimentX
http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf no 37
anonymous
  • anonymous
I guess so.. \[L(y'') = s^2Y - s(6) -0=s^2Y-6s\]What's wrong?
experimentX
  • experimentX
\[ sY'-0 +a(sY-6)-2a^2Y=0 \] this should be like \[ s^2 Y - 6s + a(sY - 0) -2a^2 Y = 0 \]
anonymous
  • anonymous
\[L(\frac{dy}{dx}) = sY - f(0)\]Right?
experimentX
  • experimentX
yep!!
experimentX
  • experimentX
woops!! sorry I misread ... i put opposite!!
experimentX
  • experimentX
\[ s^2 Y - 0 - 6 + a(sY - 6) - 2a^2 Y = 0\]
anonymous
  • anonymous
It's okay! But what's wrong with my workings?
anonymous
  • anonymous
\[L(y'') = s^2Y - s(6) -0=s^2Y-6s\]
experimentX
  • experimentX
the second step!! lol ... i again put it wrong!!
anonymous
  • anonymous
The second step is right. You just get the third immediately.. \[L(y'') = sL(y')-f'(0) = s^2 Y-sf(0) - f'(0)\] Anyway.. I still get the third step..?!
experimentX
  • experimentX
Y' ... well i thought you differentiated it.
experimentX
  • experimentX
the answer should be according to mathematica \[ \left\{\left\{y[x]\to 2 e^{-2 a x} \left(1+2 e^{3 a x}\right)\right\}\right\} \]
anonymous
  • anonymous
Sorry :(
anonymous
  • anonymous
Yes... But I can't get it :(
experimentX
  • experimentX
the final answer is right ... you get \[ F(s) = \frac{6(a+s)}{s^2 + as - 2a^2}\] looks lkie you factorized it.
experimentX
  • experimentX
now just take the inverse transform.
anonymous
  • anonymous
Factorize it -> partial fraction -> inverse Right?
experimentX
  • experimentX
yeah!! that's the way
anonymous
  • anonymous
\[F(s) = \frac{6(a+s)}{s^2 + as - 2a^2}=\frac{6(a+s)}{(a+2a)(s-a)}\]I suppose..
anonymous
  • anonymous
s+2a typo
experimentX
  • experimentX
i guess from this point on you get the answer on next step http://www.wolframalpha.com/input/?i=partial+fraction+6%28a%2Bs%29%2F%28s^2+%2B+a+s+-+2a^2%29 http://www.wolframalpha.com/input/?i=Inverse+Laplace+transform++6%28a%2Bs%29%2F%28s^2+%2B+a+s+-+2a^2%29
anonymous
  • anonymous
\[Y = \frac{6a+6s}{(s+2a)(s-a)}=\frac{A}{s+2a}+\frac{B}{s-a}\] A(s-a) + B(s+2a) =6a+6s A+B = 6 2B-A = 6 => B=4 , A=2 \[Y =\frac{2}{s+2a}+\frac{4}{s-a}\]\[y=2e^{-2at}+4e^{at}\]Why couldn't I get this answer... :\ Thanks!!

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