# Differential equation with Laplace transform

1. Feb 3, 2010

### springo

1. The problem statement, all variables and given/known data
$$t y'' + (3 t-1) y' + 3 y = 6 e^{-3t}$$
y(0) = 1
y(5) = 2

2. Relevant equations

3. The attempt at a solution
I tried applying the Laplace transform to the equation but I was having a little trouble...
L [t·y''] = -dL[y'']/ds = s2·Y' + 2·s·Y - y(0) = s2·Y' + 2·s·Y - 1
L [t·y'] = -dL[y']/ds = s·Y' + Y
Therefore:
$$s^2 Y' + 2 s Y - 1 + 3 s Y' +3 Y - s Y + 1+ 3 Y = \frac{6}{s+3}$$
$$(s^2+3s)Y'+(s+6)Y=\frac{6}{s+3}$$

And then solve the Y(s) and then use the inverse transform.
But it doesn't yield the correct result. I did the rest on the computer so I guess it's when I apply the transform where I make a mistake.