Weierstass's Excess Function - minimizing curve

Using the Weierstass condition, find the strongly minimizing curve and the value of $\displaystyle J_{min}$ for the cases

$\displaystyle \displaystyle \int_1^2 (t^3 \dot{x}^2 + 3tx^2) \ dt$

where $\displaystyle x(1) = 0, x(2) = 3$.

I've gotten to a point where I have tried to continue, but I can't seem to get anything that resembles a correct solution.

Let $\displaystyle f = t^3 \dot{x} ^2 + 3tx^2$

Using the Euler-Largrange eqn, I get

$\displaystyle 6tx - 2t^3 \ddot{x} = 0$ so $\displaystyle \ddot{x} t^2 - 3x = 0$ ...(1)

which is a Euler's type DE which can be solved by changing variable $\displaystyle t = e^s$.

Let $\displaystyle x(s) = x(t) = x(e^s)$. By the chain rule and noting that $\displaystyle \frac{dt}{ds} = e^s = t$ :

$\displaystyle \displaystyle \frac{dx}{ds} = \frac{dx}{dt} \frac{dt}{ds} = e^s \frac{dx}{dt} = t \frac{dx}{dt}$ ........(2)

and

$\displaystyle \displaystyle \frac{d^2 x}{d^2 s} = \frac{d}{ds} \big( t \frac{dx}{dt} \big) = t \frac{dx}{dt} + t \frac{d}{ds} \frac{dx}{dt} = t \frac{dx}{dt} + t^2 \frac{d^2 x}{dt^2}$ ....(3)

Subtracting (2) and (3) I get,

$\displaystyle \displaystyle \frac{d^2 x}{d^2 s} - \frac{dx}{ds} = t^2 \frac{d^2 x}{dt^2} = 3x$

implying $\displaystyle \displaystyle \frac{d^2 x}{d^2 s} - \frac{dx}{ds} - 3x = 0$

which is equivalent to equation (1).

The extremal is therefore

$\displaystyle \displaystyle x = x(t) = x(s) = Ae^{\frac{1}{2}(1 + \sqrt{13}) s} + Be^{\frac{-1}{2} (-1 + \sqrt{13})s}$

Now, when using $\displaystyle x(1) = 0, x(2) = 3$ I don't get a solution that appears to be correct.....and I don't know how to continue. Any help would be greatly appreciated.

Thank-you.