Using Laplace transform to solve ODE

[Rumpelstiltzkin]

Here's the problem:

Solve $$ty'' + 2y' + ty = 0$$
with $$lim_{t\rightarrow0+} y = 0$$
and
$$y(\pi) = 0$$

So here's my working:

$$-\frac{d}{ds}[s^2Y - sy(0) - y'(0)] + 2[sY - y(0)] - Y' = 0$$

$$-[2sY + s^2Y' - s] + 2[sY - y(0)] - Y' = 0$$

Arranging terms, I get:

$$Y' = \frac{s-2y(0)}{s^2+1}$$
and
$$-Y' = \frac{2y(0)-s}{s^2+1}$$

$$L^{-1}(-Y')= L^{-1}(\frac{2y(0)}{s^2+1}) - L^{-1}(\frac{s}{s^2+1})$$

$$ty = 2y(0)sint - cost$$

$$y = \frac{2y(0)sint}{t} - \frac{cost}{t}$$
The answer is $$y = \frac{sint}{t}$$

What's wrong?

Thanks..

 

[Nino MMP]

The problem is that you have not included the initial condition y(0) in your solution. The general solution of the differential equation is y = \frac{2y(0)sint}{t} - \frac{cost}{t}, but since y(0) = 0, the solution simplifies to y = \frac{sint}{t}.