Find a solution to $x''-2x'+tx=0$ under the assumption that $x(0)=0$. I took the Laplace of all parts to find that $$s^2-2s=\frac {X'(s)}{X(s)}$$ I am unsure of where to go from here. I am thinking that I integrate both sides, but when I look at the result of that I am not sure how that helps me. Thanks. 
 A: A more careful treatment of the equation using the Laplace transform but keeping all initial conditions non-zero (both the value of the function and its derivative are potentially non-zero) reveals that:
$$\frac{dX}{ds}=(s^2-2s)X+(2-s)x(0)-x'(0)$$
We know that $X(\infty)=0$ (we make that assumption when we integrate by parts to calculate the Laplace transform of derivatives), so we are going to integrate the equation as follows:
$$\int\limits_{\infty}^{s}\frac{d}{dw}(e^{-\frac{w^3}{3}+w^2}X(w))dw=\int\limits_{\infty}^{s}(2x(0)-x'(0)-wx(0))e^{-\frac{w^3}{3}+w^2}dw$$
The integral on the right hand side converges for all $s$ finite. Note that in the derivation of the Laplace-transformed equation it has been implicitly assumed that \textbf{both} $x(0)$ and $ x'(0)$ are equal to zero. In that case, since the equation is linear and the point $t=0$ is regular, if we assume that all initial conditions are zero, the solution has to be identically zero. This is confirmed by the Laplace transform treatment, because the RHS degenerates and we get:
$$X(s)=e^{\frac{s^3}{3}-s^2}\lim\limits_{s\rightarrow\infty}(X(s)e^{-\frac{s^3}{3}+s^2})=0$$
If you only assume that $x(0)=0$ then you get an integral representation for the solution x(t) which reads:
$$x(t)=\frac{x'(0)}{2\pi i}\int\limits_{\sigma-i\infty}^{\sigma+i\infty}ds~~e^{\frac{s^3}{3}-s^2+st}\int\limits_{s}^{\infty}e^{-\frac{w^3}{3}+w^2}dw$$
Unfortunately this expression doesn't look very useful on first sight. However if we perform a transformation to standardize the ODE and remove the first derivative term, (we set $y(t)=e^{-t}x(t)$) we obtain:
$$y''(t)+(t-1)y(t)=0$$
which looks surprisingly close to the Airy equation,and indeed, if we set $w=a(t-1)$ where $a^3=-1$ we see that:
$$y''(w)-wy(w)=0$$ 
and therefore the solution to the original ODE can be written as:
$$x(t)=e^t(C_1\text{Ai}(a(t-1))+C_2\text{Bi}(a(t-1))$$
and the most general solution satisfying x(0)=0 is:
$$x(t)=Ce^t(\text{Bi}(-a)\text{Ai}(a(t-1))-\text{Ai}(-a)\text{Bi}(a(t-1))$$.
where $a$ can be chosen to be any cubic root of unity.
This function definitely possesses a Taylor series representation which is tractable around t=0 and an integral representation like the one derived above using LT but it's definitely more difficult to connect the dots by using the aforementioned methods.
A: $$s^2-2s=\frac{X'(s)}{X(s)}$$
$$\int\left[s^2-2s\right]ds=\int\frac{X'(s)}{X(s)}ds$$
$$\int\left[s^2-2s\right]ds=\int\frac{1}{X(s)}dX(s)$$
$$\frac{s^3}{3}-s^2+C_1=\ln\left(X(s)\right)$$
$$X(s)=\exp\left[\frac{s^3}{3}-s^2+C_1\right]$$
