What went wrong? [One-dimensional-inverse-square-law]

Your manipulations are quite nice, and I enjoyed reading them, but I thought I'd point out they can be put in the language of conservation laws. You started with a differential equation

\begin{align*} \ddot x(t) = F(x(t)) && x(0) =x_0 && \dot x(0) = v_0 \end{align*}

where $F(x)$ can be thought of as a "force field" on some 1D domain in $\mathbb{R}$. Now let $V = - \int F$ be a potential function (the choice of antiderivative turns out not to affect anything). Basically, you took $\ddot x = F(x)$, wrote it as $ \dot x \ddot x = \dot x F(x) $ and used the chain rule to identify that with $\frac{d}{dt}\left( \frac{ \dot x^2}{2} \right)= - \frac{d}{dt} V(x)$ or, in other words, $$ \frac{d}{dt} \left( \frac{\dot x^2}{2} + V(x) \right) = 0$$ which says that the kinetic energy $K = \frac{\dot x^2}{2}$ and the potential energy $V(x)$ add to a constant $E$ which is fruitfully interpreted as the total energy of the system $$ \frac{\dot x^2}{2} + V(x) = E.$$ We can calculate $E$ straight away from the initial conditions: $E = \frac{1}{2} v_0^2 + V(x_0)$. Solving for the velocity, we get $$ \dot x = \pm \sqrt{2(E-V(x(t))) }$$ which is a 1st order, automomous (in particular, separable) ODE. According as the intial velocity $v_0$ is positive or negative, we can safely assume that $\dot x(t)$ will retain its sign for a short time afterwards and remove the $\pm$ sign. For instance, if we suppose $v_0 > 0$, then, for sufficiently small $t$, we apply the usual technique of separating variables: $$ t = \int_0^t \ ds = \int_0^t \frac{ \dot x(s) }{ \sqrt{ 2(E-V(x(s)))} } \ ds = \int_0^{x(t)} \frac{1}{ \sqrt{2(E-V(x))}} \ dx$$ which yields (assuming the integral can be solved) an equation relating $x(t)$ and $t$. In the case of the inverse square attractive force, we have $F(x) = \frac{-k}{x^2}$ whence $V = -\frac{k}{x}$ is a potential function so the integral becomes $$ t = \int_0^{x(t)} \frac{1}{ \sqrt{2\left( E+ \frac{k}{x} \right)}} \ dx $$ which is a bit beyond me, but not beyond most integration packages.

Lets write $$ \dot{x} = \sqrt{C-\frac{k}{x}} = \sqrt{C}\sqrt{1 - \frac{k}{Cx}}. $$ where i have absorbed the factor 2 into the constants k and C. We can solve as follows: $$ \int \frac{1}{\sqrt{1 - \frac{k}{Cx}}}dx = \sqrt{C}t + \lambda_{1} $$ If we use the transformation $u = x - \frac{k}{C}$, we transform the r.h.s as,

$$ \int \sqrt{1 + \frac{k}{Cu}}du = \sqrt{C}t + \lambda_{1}. $$

Integrating by parts with, $$ d\bar{u} = 1,\\ v = \sqrt{1 + \frac{k}{Cu}}. $$

This leads to:

$$ \int \sqrt{1 + \frac{k}{Cu}}du = u\sqrt{1 + \frac{k}{Cu}} - \int u\frac{-\frac{k}{2Cu^{2}}}{\sqrt{1 + \frac{k}{Cu}}}du,\\ =u\sqrt{1 + \frac{k}{Cu}} + \frac{1}{2}\int \frac{1}{\sqrt{1 + \frac{k}{Cu}}}\frac{k}{Cu}du $$

We can sort out the first term in terms of x already as $$ u\sqrt{1 + \frac{k}{Cu}} = \sqrt{u}\sqrt{u + \frac{k}{C}} = \sqrt{x -\frac{k}{C}}\sqrt{x},\\ =x\sqrt{1 - \frac{k}{Cx}} $$

Now the integral: $$ \frac{1}{2}\int \frac{1}{\sqrt{1 + \frac{k}{Cu}}}\frac{k}{Cu}du $$

using another transformation $w^{2} = 1 + \frac{k}{Cu} $ which has an inverse of $u = \frac{k/C}{w^{2}-1}$, we can transform the integral to,

$$ \frac{1}{2}\int \frac{w^{2} - 1}{w}\left(\frac{-2(k/C) w}{\left(w^{2}-1\right)^{2}}\right)dw = -\frac{k}{C}\int \frac{1}{w^{2}-1}dw $$

which integrates to:

$$ -\frac{k}{2C} \mathrm{log}\left(\frac{w-1}{w+1}\right) = \frac{k}{2C} \mathrm{log}\left(\frac{w+1}{w-1}\right) = \frac{k}{2C} \mathrm{log}\left(\frac{(w+1)^{2}}{w^{2} - 1}\right). $$

The log argument can be manipulated to yield: $$ \frac{(w+1)^{2}}{w^{2} - 1} = \frac{\frac{k}{Cu} + 2 + 2w}{\frac{k}{Cu}} = 1 + \frac{2Cu}{k} - 2w\frac{Cu}{k},\\ =\frac{2Cx - k + 2wCu}{k} = \frac{2Cx - k - 2\sqrt{1 + \frac{k}{Cu}}Cu}{k},\\ =\frac{2Cx - k + 2C\sqrt{u + \frac{k}{C}}\sqrt{u}}{k} = \frac{2Cx - k + 2Cx\sqrt{1 - \frac{k}{Cx}}}{k} $$

Tying everything together we find: $$ x\sqrt{1 - \frac{k}{Cx}} +\frac{k}{2C} \mathrm{log}\left(\frac{2Cx - k + 2Cx\sqrt{1 - \frac{k}{Cx}}}{k}\right)\\ = x\sqrt{1 - \frac{k}{Cx}} +\frac{k}{2C} \mathrm{log}\left(2Cx - k + 2Cx\sqrt{1 - \frac{k}{Cx}}\right) +\frac{k}{2C}\mathrm{log}\left(\frac{1}{k}\right)\\ =\sqrt{C}t + \lambda_{1} $$ The third term is a constant and can be absorbed by $\lambda_{1}$

$$ \frac{1}{\sqrt{C}}x\sqrt{C - \frac{k}{x}} +\frac{k}{2C} \mathrm{log}\left(2Cx - k + 2\sqrt{C}x\sqrt{C - \frac{k}{x}}\right) = \sqrt{C}t + \lambda_{1} $$

multiplying the $\sqrt{C}$ we finally reach

$$ \sqrt{2}x\sqrt{C - \frac{k}{x}} +\frac{k}{\sqrt{2C}} \mathrm{log}\left(2Cx - k + 2\sqrt{C}x\sqrt{C - \frac{k}{x}}\right) = \sqrt{2}Ct + \lambda_{2} $$

I final not worth respect to getting back to the original equation, you have to do implicit differentiation and re-arrange for $\frac{dx}{dt}$, thats why you had a issues when you neglect the r.h.s of the equation i.e. the t term.

Your problem seems to be much more difficult than it looks at first glance. Using a CAS, I have not been able to solve anything using the boundary conditions. Forgetting them, it looks that the solution is something such as
$$\left(\frac{x(t) \sqrt{c_1-\frac{2 k}{x(t)}}}{c_1}+\frac{k \log \left(\sqrt{c_1} x(t) \sqrt{c_1-\frac{2 k}{x(t)}}+c_1 x(t)-k\right)}{c_1^{3/2}}\right){}^2=\left(c_2+t\right){}^2$$ The boundary conditions lead to a very complex system of two nonlinear equations from which $c_1$ and $c_2$ would be expressed as functions of $x_0$ and $v_0$.

I shall try to continue but it does not look to be very promising.