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(again)Solve by Laplace transform:
\[\frac{d^2y}{dx^2} + a \frac{dy}{dx}  2a^2y=0\]
y(0)=6, y'(0) = 0
 one year ago
 one year ago
(again)Solve by Laplace transform: \[\frac{d^2y}{dx^2} + a \frac{dy}{dx}  2a^2y=0\] y(0)=6, y'(0) = 0
 one year ago
 one year ago

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RolyPolyBest ResponseYou've already chosen the best response.0
\[\frac{d^2y}{dx^2} + a \frac{dy}{dx}  2a^2y=0\]\[sY'0 +a(sY6)2a^2Y=0\]\[s(sY6) +a(sY6)2a^2Y=0\]\[(s^2+as2a^2)Y=6a+6s\]\[Y=\frac{6a+6s}{(s+2a)(sa)}\]So far so good? Or...??
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1355844666804:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
sorry ... i didn't see, looks like the transform of d^2y/dx^2 is incorrect.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf no 37
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
I guess so.. \[L(y'') = s^2Y  s(6) 0=s^2Y6s\]What's wrong?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
\[ sY'0 +a(sY6)2a^2Y=0 \] this should be like \[ s^2 Y  6s + a(sY  0) 2a^2 Y = 0 \]
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
\[L(\frac{dy}{dx}) = sY  f(0)\]Right?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
woops!! sorry I misread ... i put opposite!!
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
\[ s^2 Y  0  6 + a(sY  6)  2a^2 Y = 0\]
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
It's okay! But what's wrong with my workings?
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
\[L(y'') = s^2Y  s(6) 0=s^2Y6s\]
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
the second step!! lol ... i again put it wrong!!
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
The second step is right. You just get the third immediately.. \[L(y'') = sL(y')f'(0) = s^2 Ysf(0)  f'(0)\] Anyway.. I still get the third step..?!
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
Y' ... well i thought you differentiated it.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
the answer should be according to mathematica \[ \left\{\left\{y[x]\to 2 e^{2 a x} \left(1+2 e^{3 a x}\right)\right\}\right\} \]
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
Yes... But I can't get it :(
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
the final answer is right ... you get \[ F(s) = \frac{6(a+s)}{s^2 + as  2a^2}\] looks lkie you factorized it.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
now just take the inverse transform.
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
Factorize it > partial fraction > inverse Right?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
yeah!! that's the way
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
\[F(s) = \frac{6(a+s)}{s^2 + as  2a^2}=\frac{6(a+s)}{(a+2a)(sa)}\]I suppose..
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
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
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
\[Y = \frac{6a+6s}{(s+2a)(sa)}=\frac{A}{s+2a}+\frac{B}{sa}\] A(sa) + B(s+2a) =6a+6s A+B = 6 2BA = 6 => B=4 , A=2 \[Y =\frac{2}{s+2a}+\frac{4}{sa}\]\[y=2e^{2at}+4e^{at}\]Why couldn't I get this answer... :\ Thanks!!
 one year ago
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