Any hint at this:
$$y''\left( t\right) +\dfrac {1}{y^{2}\left( t\right) +a^{2}}=0$$
With initial conditions $y(0)=1$ and $y'(0)=0$ . I've already tried multiplying by $y$', integrating and making $$y\left( t\right) =a \tan\left( s\right)$$ but it just getting worse.
Try to get the first integral right. Multiply with $2y'$ and integrate to get $$ C=y'^2+2\int \frac{dy}{y^2+a^2}=y'^2+\frac2a\arctan\frac{y}{a} $$ Now insert initial conditions to get $C=\frac2a\arctan\frac{1}{a}$.