x(t) = [-9A/2]^(1/3) * t^(2/3)

Well, is this the only solution?

That is a very good question Wasi.

At first I was thinking exponential but by manipulating I got:

[dx/dt]^2 = -2A * d/dt(1/x)

and then guessed at a power function: x = C*t^m

So x^2 = C^2*t^(2m) and x'' = Cm(m-1) * t^(m-2)

That gave me 2m+m-2 = 0 and m = 2/3. Similar for C by equating co-efficient to A.

It reminded me of Newton's Law of Gravitation:

acceleration is prop. the inverse square of position.

x(t) = [-9A/2]^(1/3) * t^(2/3)

That is a very good question Wasi.

At first I was thinking exponential but by manipulating I got:

[dx/dt]^2 = -2A * d/dt(1/x)

and then guessed at a power function: x = C*t^m

So x^2 = C^2*t^(2m) and x'' = Cm(m-1) * t^(m-2)

That gave me 2m+m-2 = 0 and m = 2/3. Similar for C by equating co-efficient to A.

It reminded me of Newton's Law of Gravitation:

acceleration is prop. the inverse square of position.