Analytical solution of second order non linear ODE

Question

I have two coupled non linear ODE

$$ \begin {align} y_1''&=(y_1-y_2)+((y_1-y_2)^3)/6, \\ y_2''&=1+((y_1-y_2)^2)/2 \end{align} $$ How can I solve these equations analytically to get the value of $y_1$ and $y_2$?

Answer 1

There is a particular solution which satisfies $y_1-y_2=c$ (constant). Taking $d^2/dt^2$ of this condition and using the ODES gives $$ c+c^3/6=y_1''=y_2''=1+c^2/2. $$ This equation for $c$ is cubic, so there exists a real solution. Returning to $y_2''=1+c^2/2$, we conclude that $y_2=\frac{1}{2}(1+c^2/2)t^2+at+b$, and hence $y_1=y_2+c=\frac{1}{2}(1+c^2/2)t^2+at+b+c$. Here, $a$ and $b$ are arbitrary constants, while $c$ is a solution of the cubic above.