Solve the differential equation for obtaining $x$ as a relation of $t$: $\frac{d^2x}{dt^2}=\alpha\sqrt{x}$

Question

Question: Solve the differential equation for obtaining $x$ as a relation of $t$: $$\frac{d^2x}{dt^2}=\alpha\sqrt{x}$$

My attempt: $$\frac{d^2x}{dt^2}=\alpha\sqrt{x}$$ $$\Rightarrow 2\frac{dx}{dt}\frac{d^2x}{dt^2}=\alpha\sqrt{x}\cdot 2\frac{dx}{dt}$$ $$\Rightarrow \frac{d}{dt}\left[\left(\frac{dx}{dt}\right)^2\right]=\frac{4}{3}\alpha\cdot \frac{d}{dt}\left(x^\frac{3}{2}\right)$$ $$\Rightarrow \left(\frac{dx}{dt}\right)^2=\frac{4}{3}\alpha x^\frac{3}{2}+c_1$$

Now we have $$\frac{dx}{dt}=\sqrt{\frac{4}{3}\alpha x^\frac{3}{2}+c_1}=\sqrt{k x^\frac{3}{2}+c_1}$$ $$\Rightarrow \frac{dx}{\sqrt{k x^\frac{3}{2}+c_1}}=dt$$

Can anyone suggest how to proceed? Any substitutions?

Answer 1

HINT:

$$x''(t)=\alpha\sqrt{x(t)}\Longleftrightarrow$$ $$\frac{\text{d}^2x(t)}{\text{d}t^2}=\alpha\sqrt{x(t)}\Longleftrightarrow$$ $$\frac{\text{d}^2x(t)}{\text{d}t^2}\cdot\frac{\text{d}x(t)}{\text{d}t}=\alpha\sqrt{x(t)}\cdot\frac{\text{d}x(t)}{\text{d}t}\Longleftrightarrow$$ $$\int\space\frac{\text{d}^2x(t)}{\text{d}t^2}\cdot\frac{\text{d}x(t)}{\text{d}t}\space\text{d}t=\int\space\alpha\sqrt{x(t)}\cdot\frac{\text{d}x(t)}{\text{d}t}\space\text{d}t\Longleftrightarrow$$ $$\frac{1}{2}\left(\frac{\text{d}x(t)}{\text{d}t}\right)^2=\frac{2\alpha x(t)^{\frac{3}{2}}}{3}+\text{C}\Longleftrightarrow$$ $$\left(\frac{\text{d}x(t)}{\text{d}t}\right)^2=\frac{4\alpha x(t)^{\frac{3}{2}}}{3}+2\text{C}\Longleftrightarrow$$ $$\frac{\text{d}x(t)}{\text{d}t}=\pm\sqrt{\frac{4\alpha x(t)^{\frac{3}{2}}}{3}+2\text{C}}\Longleftrightarrow$$ $$\frac{\frac{\text{d}x(t)}{\text{d}t}}{\sqrt{\frac{4\alpha x(t)^{\frac{3}{2}}}{3}+2\text{C}}}=\pm 1\Longleftrightarrow$$ $$\int\space\frac{\frac{\text{d}x(t)}{\text{d}t}}{\sqrt{\frac{4\alpha x(t)^{\frac{3}{2}}}{3}+2\text{C}}}\space\text{d}t=\pm\int\space 1\space\text{d}t\Longleftrightarrow$$ $$\frac{_2\text{F}_1\left(\frac{1}{2},\frac{2}{3};\frac{5}{3};-\frac{2\alpha x(t)^{\frac{3}{2}}}{3\text{C}}\right)\cdot x(t)\sqrt{\frac{2\alpha x(t)^{\frac{3}{2}}}{\text{C}}+3}}{\sqrt{6\text{C}+4\alpha x(t)^{\frac{3}{2}}}}=\pm t+\text{F}$$

With $\text{C}$ and $\text{F}$ is an arbitrary constants