Laplace Transform With Initial Values

[diffeqnoob]

Okay, I know this is alot... but I am stuck, so here goes...

Use the method of Laplace transform to solve the initial value problem

y''+3ty'-6y=0, y(0) = 1, y'(0) = 0
L\{y'' + 3ty' - 6y\} = L\{0\}
s^{2}Y(s) - sy(0) - y'(0) + 3L\{ty'\} - 6Y(s) = 0
s^{2}Y(s) - s(1) - 0 - \frac{d}{ds}\left(3 L\{ty'\}\right) -6Y(s) = 0

Now to resolve the - \frac{d}{ds}\left(3 L\{ty'\}\right)
= - \frac{d}{ds}\left(3 L\{ty'\}\right)

= - \frac{d}{ds}3 \left(sY(s) - y(0)\right)

= -3sY'(s) - 3Y(s)


Plugging it back into the eq we now have

s^{2}Y(s) - s - 3sY'(s) - 3Y(s) - 6Y(s) = 0

-3sY'(s) + (s^{2}-9)Y(s) - s = 0

Y'(s) + \left(-\frac{s}{3} + \frac{3}{s}\right)Y(s) = -\frac{1}{3}

\mu = e^{\int\left(-\frac{s}{3} + \frac{3}{s}\right)ds}

\mu = e^{\left(-\frac{s^{2}}{6} + ln(s^{3})\right)}

\mu = s^{3}e^{-\left(\frac s^{2}{6}\right)}

\int\left(\frac{d}{ds}(s^{3}e^{-\left(\frac s^{2}{6}\right)}Y(s)\right) = \int-\left(\frac{1}{3}\right)s^{3}e^{-\left(\frac s^{2}{6}\right)} ds

s^{3}e^{-\left(\frac s^{2}{6}\right)}Y(s) = \int-\left(\frac{1}{3}\right)s^{3}e^{-\left(\frac s^{2}{6}\right)} ds


RIGHT SIDE
=\left(\frac{1}{3}\right)(-3(s^{2}+6)e^{-\left(\frac{s^2}{6}\right)

=(s^2+6)e^{-\left(\frac{s^2}{6}\right) + A

Y(s)=\frac{(s^2+6)}{s^{3}} + \frac{A e^ \frac{s^2}{6}}{s^{3}}

Limit as s \rightarrow \infty Y(s) = 0 therefore A = 0

Y(s) = \frac{s^2+6}{s^3}



Break down the Inverse Laplace
L^{-1}\{\frac{s^2+6}{s^3}\}

=L^{-1}\{\frac{s^2}{s^3}\} + L^{-1}{\frac{6}{s^3}\}

=L^{-1}\{\frac{1}{s}\} + L^{-1}{\frac{6}{s^3}\}

= 1 + ?


This is where I get lost... I don't know how to do the other side... Please help.

 

[diffeqnoob]

Why does my LaTex look all funny?