Homework Help: Laplace Transformation Question

1. Oct 8, 2011

Applejacks

1. The problem statement, all variables and given/known data

Use Laplace transformations to solve:
Y''+tY'+2Y=0, Y(0)=0, Y'(0)=1

2. Relevant equations

Y''==>$s^{2}$y-sY(0)-Y'(0)=$s^{2}$y-1
tY'==>-(y+sy')
2Y==>2y

3. The attempt at a solution

The question felt pretty straightforward till I hit a rock.

$s^{2}$y-1+y-sy'=0

y'=$\frac{1}{s}$(y($s^{2}$+1)-1)

My problem is that I cant integrate this because of the -1. What to do? Am I on the right track?

(First time using these forums and symbols so give me a sec to fix them up. Not sure why the S is coming out weird.)

Last edited: Oct 8, 2011
2. Oct 8, 2011

Kurt Peek

Hi Applejacks,

I verified your calculations, and I think you are on the right track. The $-1$ does indeed preclude the use of separation of variables to solve the ODE for $y(s)$. If I put that ODE into Maple I get a complicated expression involving the error function, so it does not look like there's an easy way; maybe you could look up the integral in, for example, Abramowitz and Stegun.

Cheers,
Kurt

P.S. According to Maple the solution of your ODE with boundary conditions is $$Y(t)=t \exp{(-t^2/2)}.$$

3. Oct 8, 2011

vela

Staff Emeritus
Rearranging terms a little, you get
$$sy' - (s^2+1)y = -1$$which is a first-order differential equation. I'd try solving it using an integrating factor.

4. Oct 8, 2011

Kurt Peek

Hi,

Indeed, vela, I had forgotten about that technique. Upon investigation I find that it does not make the problem much easier, however. By using the integrating factor technique, I recast the first-order ODE for $y(s)$ into the following form: $$\frac{d}{ds}\left(\frac{1}{s}\exp{\left(-\frac{s^2}{2}\right)}y(s)\right)=-\frac{1}{s}\exp{\left(-\frac{s^2}{2}\right)}.$$ Integrating this equation requires integration of the Gaussian function on the right hand side, which leads to an expression involving an error function (in agreement with the Maple output I mentioned earlier).

Cheers,
Kurt

P.S. That the expression for $y(s)$ contains an error function can also be seen by taking the Laplace transform of the known solution $Y(t)=t\exp(-t^2/2)$ that I posted earlier.