Velocity/Acceleration Proof

I want to assume that a body moving with velocity $v$ encounters resistance of the form $\frac{dv}{dt} = -kv^{\frac{3}{2}}$ and show that $v(t) = \frac{4v_{0}}{{(kt\sqrt{v_{0}}+2})^2}$ and that $x(t) = x_{0} + \frac{2}{k}\sqrt{v_{0}}(1-\frac{2}{kt\sqrt{v_{0}}+2})$

Let me first explain what the given equation represents. It states the acceleration of the body is the rate of the velocity per rate of time and that is proportional to $v^{\frac{3}{2}}$. The $-k$ on the right hand side of the differential equation means that there is some constant $k$ that represents a “resisting” force that is opposite to the direction of the velocity. Now to solve the differential equation.

Since the equation is separable it can broken apart into the following equation $v^{\frac{-3}{2}} dv = -k dt$. Since I now have both an integrand and a differential element, I can sum up all of the differential elements by integrating which will look like $\int v^{\frac{-3}{2}} dv = \int -k dt$. After integration is complete, I obtain the equation, $-2v^{\frac{-1}{2}} = -kt - C$. After dividing the equation by $-2$ and cross multiplying, I obtain the formula $v^{\frac{1}{2}} = \frac{2}{kt + C}$. Now, I can square both sides of the equation and solve for $v(t)$. This formula will look like $v(t) = \frac{4}{(kt+C)^2}$.  At time $t=0$, the velocity is equal to the initial velocity, $v_{0}$. When I plug in these conditions to the equation, and take the positive root since velocity is positive, I can solve for the constant $C$, to find that $C = \frac{2}{\sqrt{v_{0}}}$. I can then plug this equation into my derived velocity equation to obtain $v(t) = \frac{4}{(kt+\frac{2}{\sqrt{v_{0}}})^2}$ After some minor algebraic modifications, the equation becomes $v(t) = \frac{4v_{0}}{{(kt\sqrt{v_{0}}+2})^2}$ which is what I was seeking to show.

From this, I can write the final equation above as $\frac{dx}{dt} = \frac{4v_{0}}{(kt\sqrt{v_{0}}+2)^2}$. Like the given differential equation, this equation can be separated and integrated which will look as follows$\frac{1}{4v_{0}} \int 1 dx = \int (kt\sqrt{v_{0}} + 2)^{-2} dt$. Using $u$ substitution, I can let $u = (kt\sqrt{v_{0}}+2$. This then implies that $du = k\sqrt{v_{0}} dt$. After making the substitution for $u$ I must multiply the integral by the constant $\frac{k\sqrt{v_{0}}}{k\sqrt{v_{0}}}$, and bring the constant in the numerator into the integral I can now carry out the $u$ substitution for $du$ and integrate which will give me the formula $\frac{x}{4v_{0}} = \frac{-u^{-1}}{k\sqrt{v_{0}}}+C$. To find the constant $C$ I notice that at time $t = 0$, the position is $x_{0}$. After making this substitution for $t$ and $x$ I find that $C=\frac{x_{0}k+2\sqrt{v_{0}}}{4kv_{0}}$. I can then plug this into the equation that I plugged the initial conditions into, and this will give me $\frac{x}{4v_{0}} = \frac{-({kt\sqrt{v_{0}}+2)}^{-1}}{k\sqrt{v_{0}}} +\frac{x_{0}k+2\sqrt{v_{0}}}{4kv_{0}}$. Finally I can find a common denominator and cancel common terms to obtain the the equation $x(t) = \frac{-4\sqrt{v_{0}}}{k^{2}t\sqrt{v_{0}}+2k} + x_{0} + \frac{2\sqrt{v_{0}}}{k}$. Then I can factor out $\frac{2\sqrt{v_{0}}}{k}$ from two of the terms to obtain the final formula $x_{0} + \frac{2\sqrt{v_{0}}}{k} (1- \frac{2}{kt\sqrt{v_{0}}+2})$, which is what I was asked to show. Now, if i take $\displaystyle \lim_{t\to\infty}$ I obtain the function $x(t) = x_{0} + \frac{2\sqrt{v_{9}}}{k}$ which implies that the object travels only a finite distance before coming to a stop.