Problem : $y''+8ty'-16y = 3 , y(0) = y'(0) = 0 $
I am supposed to use the series method to get F(s) , then get the inverse laplace transform to get f(t).
I got the Laplace transform : $(s^3 - 24s) F(s) - 8s^2F'(s) = 3 $
then I started using the series method and ended up with $\sum\limits_{n=3}^\infty [C_{n-3} -16C_{n-1} - 8n C_{n-1} ]s^n = 3 $
and I am not sure if that's right since the coefficient of $x^{0}$ is clearly not 0 .
Integrating by parts twice we have: \begin{equation} \int\limits_0^\infty y^{''}(t) e^{-s t} dt = -y^{'}(0) - s y(0) + s^2 F(s) \end{equation} Likewise: \begin{equation} \int\limits_0^\infty y^{'}(t) e^{-s t} dt = -y(0) + s F(s) \end{equation} finally: \begin{equation} \int\limits_0^\infty t y^{'}(t) e^{-s t} dt = \left. y(t) t e^{-s t} \right|_0^\infty - \int\limits_0^\infty y(t) \left( 1 e^{-s t} + t (-s) e^{-s t}\right) dt = - F(s) - s F^{'}(s) \end{equation} Bringing all terms together we get: \begin{equation} \left(s^2 - 24 \right) F(s) - 8 s F^{'}(s) = \frac{3}{s} \end{equation}
Now, we make the usual ansatz $F(s) = \sum\limits_{n=0}^\infty a_n s^{n+r} $. Inserting this into the equation we get:
\begin{equation} \sum\limits_{n=2}^\infty \left(a_{n-2} - a_n (24 + 8 (n+r))\right) s^{n+r} = 0 \end{equation} and $-24 a_0 s^r - 8 a_0 r s^r = 3/s$ and $-24 a_1 s^{r+1} - 8 a_1 (r+1) s^{r+1}= 0$. Thus $r=-1$, $a_0 = -3/16$, $a_1=0$ and $a_{n-2} = a_n (16 +8 n)$. This leads to the following solution:
\begin{equation} F(s) = \frac{3\left(1 - e^{\frac{s^2}{16}} \right)}{s^3} \end{equation}
However, instead of assuming the power series expansion we could just assume $F(s) = A s^r$. Inserting this into the equation for the Laplace transform leads to another solution:
\begin{equation} F(s) = \frac{3}{s^3} \end{equation}
which is readily inverted and gives
\begin{equation} f(t) = \frac{3 t^2}{2} \end{equation}
By inserting the last equation into the original ODE we verify that this is the correct solution.