I was given the following problem:
Given $$C\ge2, v_0=\sqrt C \ ,\ x_0=C\ ,\ f(x)=x''=\frac{-C}{x^2}$$ find $t>0$ satisfying $x(t)=2$.
What I did:
Using the energy equation, $\frac{x'(t)^2}{2}-\int_{x_0}^{x(s)}f(s)ds=\frac{v_0^2}{2}$, we find $x_{max}$, which implies $v=0\iff x'(t)=0$: $$0+\int_{C}^{x_{max}}\frac{C}{s^2}ds=\frac{v_0^2}{2}\\\Rightarrow x_{max}=\frac{C}{1-\frac{v_0^2}{2}}$$
And using the energy equation again we find $t^*$: $$\frac{x'(t)^2}{2}+\int_{C}^{x_{t}}\frac{C}{s^2}ds=\frac{v_0^2}{2}\iff x'(t)^2+2C(-\frac{1}{x}+1)=v_0^2 \\\iff x'(t)=\pm\sqrt{v_0^2-2C+\frac{2C}{x}} \\\Rightarrow\int_{C}^{2}\frac{dx}{\pm\sqrt{v_0^2-2C+\frac{2C}{x}}}=\int_{0}^{t^*}dt=t^*$$
Following this point I think the solution with the minus sign in the fraction is wrong, as I was asked to give a positive answer. The problem I have, is the instructions say I should encounter two equations of the form: $$\int\frac{dx}{\sqrt{\frac{a}{x}+b}}$$
$a>0$ in both equations, but $b>0$ in one equation and $b<0$ in the other.
My solution only encounters the $b<0$ version.
I get the conserved energy term as $$ E(x,v)=\frac{v^2}2-\frac{C}x $$ so that $E_0=E(x_0,v_0)=\frac C2-1>0$ and the resulting first order equation $$ x'=-\sqrt{\frac{2C}x+2E_0}=-\sqrt{\frac{2C}x+C-2} $$ (want to fall down from $x_0>2$ to $x=2$) which corresponds to your case $a>0$, $b>0$.
However, there is no positive $x_{\max}>x_0$ to be found.