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?