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

This Question is Closed

anonymous
 3 years ago
Best 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...??

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1355844666804:dw

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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[L(\frac{dy}{dx}) = sY  f(0)\]Right?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1woops!! sorry I misread ... i put opposite!!

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1\[ s^2 Y  0  6 + a(sY  6)  2a^2 Y = 0\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It's okay! But what's wrong with my workings?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[L(y'') = s^2Y  s(6) 0=s^2Y6s\]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1the second step!! lol ... i again put it wrong!!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The 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..?!

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1Y' ... well i thought you differentiated it.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1the 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
 3 years ago
Best ResponseYou've already chosen the best response.0Yes... But I can't get it :(

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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1now just take the inverse transform.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Factorize it > partial fraction > inverse Right?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1yeah!! that's the way

anonymous
 3 years ago
Best 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..

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1i 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
 3 years ago
Best 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!!
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