Basic Kinematics?

We have the differential equation $\ddot{x}(t) = t \sqrt{x(t)}$ . This is both nonlinear and second-order, so it looks like solving it would be a lost cause. However, we can try homogenizing this equation (this won't usually work for differential equations with more than two terms). This essentially means assuming the $x(t)$ is a polynomial and requiring the degree of both sides to be equal. Say that $x(t)$ is given by a polynomial of some degree, $n$ , in $t$ . Since differentiating twice lowers the degree by $2$ , we have: $n-2 = n/2 + 1$ Solving gives us $n=6$ . Taking a minimal polynomial $x(t) = ct^6$ and plugging it into the differential equation yields: $30ct^4 = \sqrt{c}t^4$ $\sqrt{c} = \frac{1}{30}$ $c = \frac{1}{900}$ So we now know that $x(t) = \frac{1}{900}t^6$ is a solution to this differential equation. Finding its velocity leads to $\dot{x}(t) = \frac{6}{900}t^5 = \frac{1}{150}t^5$ FInally, $\dot{x}(1) = \frac{1}{150}$

Looking for a solution of the form $x = At^\alpha$ , we require that $\alpha(\alpha-1)At^{\alpha-2} \; = \; \sqrt{A}t^{\frac12\alpha+1} \hspace{2cm} t \ge 0$ so we need $\alpha(\alpha-1)\sqrt{A} = 1$ and $\alpha-2=\tfrac12\alpha+1$ . Thus $\alpha=6$ and $A = \tfrac{1}{900}$ , so that $x = \tfrac{1}{900}t^6$ and the speed $v = \tfrac{1}{150}t^5$ . At time $t=1$ , the speed is $\boxed{\tfrac{1}{150}}$ m/s.

However, another perfectly good solution of the problem is $x \equiv 0$ , so it is not entirely clear why the first solution is guaranteed!