## RolyPoly 2 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

1. RolyPoly

$\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...??

2. experimentX

|dw:1355844666804:dw|

3. RolyPoly

y(0)=6, y'(0) = 0

4. experimentX

sorry ... i didn't see, looks like the transform of d^2y/dx^2 is incorrect.

5. experimentX
6. RolyPoly

I guess so.. $L(y'') = s^2Y - s(6) -0=s^2Y-6s$What's wrong?

7. experimentX

$sY'-0 +a(sY-6)-2a^2Y=0$ this should be like $s^2 Y - 6s + a(sY - 0) -2a^2 Y = 0$

8. RolyPoly

$L(\frac{dy}{dx}) = sY - f(0)$Right?

9. experimentX

yep!!

10. experimentX

woops!! sorry I misread ... i put opposite!!

11. experimentX

$s^2 Y - 0 - 6 + a(sY - 6) - 2a^2 Y = 0$

12. RolyPoly

It's okay! But what's wrong with my workings?

13. RolyPoly

$L(y'') = s^2Y - s(6) -0=s^2Y-6s$

14. experimentX

the second step!! lol ... i again put it wrong!!

15. RolyPoly

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..?!

16. experimentX

Y' ... well i thought you differentiated it.

17. experimentX

the answer should be according to mathematica $\left\{\left\{y[x]\to 2 e^{-2 a x} \left(1+2 e^{3 a x}\right)\right\}\right\}$

18. RolyPoly

Sorry :(

19. RolyPoly

Yes... But I can't get it :(

20. experimentX

the final answer is right ... you get $F(s) = \frac{6(a+s)}{s^2 + as - 2a^2}$ looks lkie you factorized it.

21. experimentX

now just take the inverse transform.

22. RolyPoly

Factorize it -> partial fraction -> inverse Right?

23. experimentX

yeah!! that's the way

24. RolyPoly

$F(s) = \frac{6(a+s)}{s^2 + as - 2a^2}=\frac{6(a+s)}{(a+2a)(s-a)}$I suppose..

25. RolyPoly

s+2a typo

26. RolyPoly

$Y = \frac{6a+6s}{(s+2a)(s-a)}=\frac{A}{s+2a}+\frac{B}{s-a}$ 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!!