Boundary condition ODE Solving the non-linear ODE $\gamma^{\prime \prime} = \gamma \gamma'$ gives the solution
\begin{align*}
    \gamma(t) = C_1 \tan{\bigg(\frac{C_1 t + C_1C_2}{2}\bigg)}
\end{align*}
my two boundary conditions are
\begin{align*}
    \gamma(0) &=  x\\
    \gamma(1) &= y
\end{align*}
I am trying to find $C_1(x,y)$ and $C_2(x,y)$ explicitly. After trying to play around with trig identities, and re-writing tan as a product of exponentials I just cant figure out if it's possible.
I tried simplifying by setting $x=0$, and the furthest I get is that $C_2=0$ and $C_1 \tan(C_1/2) = y$. However, even in this simplified case there doesn't seem like there is a way to represent $C_1$ explicitly in terms of $y$.
 A: $$y''(t)=y'(t)y(t)$$
It's obvious that WA 's answer is coorect
$$2y''=2y'y$$
First integration gives us:
$$2y'=y^2+C_1$$
This DE is separable:
$$\int \dfrac {dy}{y^2+C_1}=\dfrac 12 t+C_2$$
For $C_1>0$ use $arctan$ function to integrate LHS.
$$ \arctan \left (\dfrac {y}{\sqrt {C_1}} \right )=\sqrt {C_1}(\dfrac 12t+C_2)$$
$$\implies y(t)=\sqrt {C_1} \tan (\sqrt {C_1}(\dfrac 12t+C_2))$$
More simply $(C_1>0)$:
$$\implies y(t)={2C_1} \tan \left (  {tC_1}+C_2 \right)$$
For $C_1=0$ it's easy to integrate the DE. Otherwise $(C_1<0)$ use partial fraction decomposition and $ln$ functions to integrate LHS.
What is not clear is the boundary condtions that depends on variable $x,y$. The $C$ are constants not functions of $x,y$
A: $$Y''(x)=Y(t)Y'(t) \implies Y''=\frac{1}{2} \frac{d}{dt}(Y^2) \implies \int Y'' dt=\frac{1}{2} Y^2  \implies  Y'= \frac{Y^2}{2}+A^2/2$$
$$\implies \int \frac{2dY}{Y^2+A^2}=\int \frac{dt}{2}+C \implies \frac{1}{A} \tan^{-1}(Y/A)=t/2+c \implies Y=A \tan(At/2+B)$$ Given that $Y(0)=x \implies A \tan B=x$.
Given $Y(1)=y \implies A \tan(A/2+B)=y$
Finally, we get $$y=A \left( \frac{A \tan(A/2)+x}{A-x\tan(A/2)} \right).$$
A: I prefer to add another answer,just dedicated to the solution of equation $(3)$ of my previous answer and few othe aspects.
We had
$$\frac{c_1 \left(c_1 \tan \left(\frac{c_1}{2}\right)+x\right)}{c_1-x \tan
   \left(\frac{c_1}{2}\right)}=y \tag 3$$ which can rewrite
$$f(c_1)=a\,c_1 \cos \left(\frac{c_1}{2}\right)
   +\sin \left(\frac{c_1}{2}\right) \left(c_1^2+b\right)=0$$ where $a=x-y$ and $b=xy$.
which, from a numerical point of view, is much better conditioned than $(3)$ since no more discontinuities.
Going to expansions around $c_1=0$, we can then write
$$\frac{f(c_1)}c=\sum_{n=0}^\infty (-1)^n \frac{ (2 a+b)+4( a-2) n-16 n^2}{2^{2 n+1}\,(2 n+1)!}\,c_1^{2n}$$ which, if truncated to $O\left(c_1^{2p+2}\right)$, reduce the problem to a polynomial of degree $p$ in $c_1^2$.
For the worked example $(x=1,y=3)$ corresponding to $(a=-2,b=3)$, the smallest positive solutions are
$$\left(
\begin{array}{cc}
p & c_1 \\
 1 &  0.8528028654 \\
 2 &  0.8647660893 \\
 3 &  0.8646548281 \\
 4 &  0.8646553082\\
 5 &  0.8646553069 \\
 6 &  0.8646553069
\end{array}
\right)$$
Let
$$A_k=(-1)^n \frac{ (2 a+b)+4( a-2) n-16 n^2}{2^{2 n+1}\,(2 n+1)!}$$ and the  $[2,6]$ Padé approximant gives
$$c_1^2=\frac{A_0 \left(A_1^3-2 A_0 A_1 A_2+A_0^2 A_3\right)}{A_1^4-3 A_0 
   A_1^2A_2+2 A_0^2 A_3 A_1+ A_0^2A_2^2-A_0^3 A_4}$$
