I can't solve a problem:
$ A= 0.5 (ms)^{-1}$, $ x_0 = 0.5 m $, $v(t)= A \cdot x^2 $, I have to compute the position at $t=3$ ($x_0$ is the initial position).
So my guess is that I should be able to compute the $x(t)$ formula by integrating $v(t)$:
$$ \int^t_0 A \cdot x^2 dt = x_0 + A \cdot x^2 \cdot t $$
So I get:
$$ x^2 \cdot At -x + x_0 = 0 $$
Which is a 2nd grade equation with a negative discriminant:
$$ \Delta(x) = (-1)^2 - 4 \cdot Atx_0 = 1 -4 \cdot 0.5 \cdot 3 \cdot 0.5 = 1-3=-2$$
My book includes just the solution, but it doesn't say how to get it. The solution is:
$$ x(t) = \frac{x_0}{1 - x_0At} $$
If I study it I get:
$$ x = \frac{x_0}{1 - x_0At} $$ $$ x \cdot \big( 1 - x_0At \big) = x_0 $$ $$ xx_0 \cdot At -x + x_0 =0 $$
Which is different from the one I got ($x^2 \cdot At -x + x_0 = 0$).