How can I solve $y''=\frac{a}{y^2}$ where a is a (positive) constant?

Question

Actually, I found out a way to solve that, but I can't get rid of complex numbers. And it does not make sense when it comes to complex numbers as the original question that involves this differential equation has nothing to do with complex numbers, as this is from a physics question.

Answer 1

HINT:

$$y''(t)=\frac{a}{y(t)^2}\Longleftrightarrow$$ $$y''(t)y'(t)=\frac{ay'(t)}{y(t)^2}\Longleftrightarrow$$ $$\int y''(t)y'(t)\space\text{d}t=\int\frac{ay'(t)}{y(t)^2}\space\text{d}t\Longleftrightarrow$$ $$\frac{y'(t)^2}{2}=\text{C}_1-\frac{a}{y(t)}\Longleftrightarrow$$ $$y'(t)^2=2\text{C}_1-\frac{2a}{y(t)}\Longleftrightarrow$$ $$y'(t)=\pm\sqrt{\text{C}_1-\frac{2a}{y(t)}}\Longleftrightarrow$$ $$\frac{y'(t)}{\sqrt{\text{C}_1-\frac{2a}{y(t)}}}=\pm1\Longleftrightarrow$$ $$\int\frac{y'(t)}{\sqrt{\text{C}_1-\frac{2a}{y(t)}}}\space\text{d}t=\int\pm1\space\text{d}t\Longleftrightarrow$$ $$\int\frac{y'(t)}{\sqrt{\frac{\text{C}y(t)-2a}{y(t)}}}\space\text{d}t=\text{C}_2\pm t$$

Now, for the integral you can substitute $u=\text{C}y(t)-2a$ and $\text{d}u=\text{C}y'(t)\space\text{d}t$.

We get:

$$\int\frac{y'(t)}{\sqrt{\frac{\text{C}y(t)-2a}{y(t)}}}\space\text{d}t=\frac{1}{\text{C}}\int\frac{1}{\sqrt{\frac{\text{C}u}{u+2a}}}\space\text{d}u$$