How to solve this quadratic nonhomogenius differential equation I have this differential equation
$r'(t) = \sqrt{1-\left(\dfrac{a}{1+a^2t^2}\right)^2 r(t)^2}$ where $r(0)=0, r'(0)=1$
I have no clue how to solve it, my usual techniques fail (such as separation of variables, direct integration, and a lot of things that don't apply because it is nonhomogenius and nonlinear). I also tried to find some substitutions to separate the variables without success.
With the rescaling $r\rightarrow r/a,t\rightarrow t/a$ as suggested by JohnBarber and a subsitution $t=\tan(u)$ (where the time domain $[0,\infty)$ maps to $[0,\pi/2)$) this can be simplified to
$r'(u)^2+r(u)^2=\dfrac{1}{\cos(u)^4}, r(0)=0, r'(0)=1$
Any suggestions would be highly appreciated.
As it was asked in the comments: the background is the following geometrical problem:
A point $A$ moves along a the line $y=1/2$ with position $x(t)=a⋅t$ with constant velocity $a>0$. A second point $B$ starting at the origin $O$ tries to "stick to" the (moving) line $AO$ while using its remaining velocity component of total velocity $|v|=1$ to move away from $A$. The movement of $B$ in polar coordinates $(r,\varphi)$ is such that $\varphi(t)$ follows trivially from the problem while $r(t)$ satisfies the above equation. More detailed, one derives that $\varphi'(t)=2a/(1+(2at)^2)$ and plugs it into $|v|=1$ which leads to the above equation.

The geometric view suggests that with inreasing distance $r$ there might be a point in time where $B$ cannot compensate the rotational motion of $A$ anymore which is when the term under the squareroot becomes negative.
 A: Starting from
$$(1+t^{2})^{2}(1-r'(t)^{2})-r(t)^{2}=0$$
postulate a power series with integer powers:
$$r=t+a_{2}t^{2}+a_{3}t^{3}+\cdots \text{(the first term deduced from i.c.)}$$
$$r'=1+2a_{2}t+3a_{3}t^{2}+\cdots$$
Using a CAS for I get:
$$-4ta_{2}+t^{2}\left(-4a_{2}^{2}-6a_{3}-1\right)+t^{3}\left(-12a_{2}a_{3}-10a_{2}-8a_{4}\right)+t^{4}\left(-9a_{2}^{2}-16a_{2}a_{4}-9a_{3}^{2}-14a_{3}-10a_{5}\right)+ 
t^{5}\left(-26a_{2}a_{3}-20a_{2}a_{5}-4a_{2}-24a_{3}a_{4}-18a_{4}-12a_{6}\right)+O\left(t^{6}\right) =0$$
Solving term by term
$$ \begin{eqnarray*}
a_{2} & = & 0\\
-4a_{2}^{2}-6a_{3}-1 & = & 0\\
-12a_{2}a_{3}-10a_{2}-8a_{4} & = & 0\\
-9a_{2}^{2}-16a_{2}a_{4}-9a_{3}^{2}-14a_{3}-10a_{5} & = & 0\\
-26a_{2}a_{3}-20a_{2}a_{5}-4a_{2}-24a_{3}a_{4}-18a_{4}-12a_{6} & = & 0
\end{eqnarray*}
$$
by inspection $a_{2},a_{4},a_{6}=0$:
$$\begin{eqnarray*}
-6a_{3}-1 & = & 0\\
-9a_{3}^{2}-14a_{3}-10a_{5} & = & 0
\end{eqnarray*}
$$
so $r=t-\frac{1}{6}t^{3}+\frac{5}{24}t^{5}+\cdots$
It seems like $r$ may be odd, but I haven't managed to prove this.
