I need to solve the following ODE:
$$
Ay''-By^3+Cy=0
$$
where $A,B,C \in \mathbb{R}_{+}$ (i.e.- positive constants).
After reading a little bit about the properties of the Jacobi sn function, I think the solution should contain sn in it, but have no idea how to arrive to it.
Will you please help me understand how to formally obtain the solution for this ode?
In addition, is there any book that contains solved examples for odes with Jacobi functions as solutions?
Thanks !
Best Answer
$$ Ay''-By^3+Cy=0 $$ This is an autonomous ODE. The method to solve it is : $$ 2A^2y''y'-2ABy^3y'+2ACyy'=0\quad\to\quad A^2(y')^2-\frac{AB}{2}y^4+ACy^2=c_1 $$ $$ Ay'=\pm\sqrt{\frac{AB}{2}y^4-ACy^2+c_1} $$ $$ dx=A\frac{dy}{\sqrt{\frac{AB}{2}y^4-ACy^2+c_1}}\quad\to\quad x=A\int \frac{dy}{\sqrt{\frac{AB}{2}y^4-ACy^2+c_1}} $$ $$ x=\sqrt{\frac{2A}{B}}\int \frac{dy}{\sqrt{(y^2-y_1^2)(y^2-y_2^2)}}= \sqrt{\frac{2A}{B}} \frac{1}{y_2}\:\text{F}\bigg(\sin^{-1}(\frac{y}{y_1})\:,\:\frac{y_1}{y_2} \bigg) $$ where $y_1^2$ and $y_2^2$ are the roots of $\quad \frac{AB}{2}Y^2-ACY+c_1=0$
F$(\:\:,\:\:)$ is the elliptic integral of first kind.
with the related inverse function : $$ y=y_1\:\text{sn}\left(\sqrt{\frac{B}{2A}} \:y_2\: x \:\:,\:\:\frac{y_1}{y_2} \right) $$ sn$(\:\:,\:\:)$ is the Jacobi elliptic sn function.