Binomial differential equation

In mathematics, the binomial differential equation is an ordinary differential equation of the form $$\left( y' \right)^m = f(x,y),$$ where $$m$$ is a natural number and $$f(x,y)$$ is a polynomial that is analytic in both variables.

Solution
Let $$ P(x,y) = (x + y)^k $$ be a polynomial of two variables of order $$k$$, where $$k$$ is a natural number. By the binomial formula,
 * $$ P(x,y) = \sum\limits_{j = 0}^k { \binom{k}{j} x^j y^{k - j} } $$.

The binomial differential equation becomes $(y')^m = (x + y)^k$. Substituting $$v = x + y$$ and its derivative $$v' = 1 + y'$$ gives $(v'-1)^m = v^k$, which can be written $\tfrac{dv}{dx} = 1 + v^{\tfrac{k}{m}}$ , which is a separable ordinary differential equation. Solving gives

\begin{array}{lrl} & \frac{dv}{dx} &= 1 + v^{\tfrac{k}{m}} \\ \Rightarrow & \frac{dv}{1 + v^{\tfrac{k}{m}}} &= dx \\ \Rightarrow & \int {\frac{dv}{1 + v^{\tfrac{k}{m}}}} &= x + C \end{array} $$

Special cases

 * If $$m=k$$, this gives the differential equation $$v' - 1 = v$$ and the solution is $$y\left( x \right) = Ce^x - x - 1$$, where $$C$$ is a constant.
 * If $$m|k$$ (that is, $$m$$ is a divisor of $$k$$), then the solution has the form $\int {\frac} = x + C$ . In the tables book Gradshteyn and Ryzhik, this form decomposes as:



\int {\frac} = \left\{ \begin{array}{ll} - \frac{2}{n}\sum\limits_{i = 0}^{{\textstyle{n \over 2}} - 1} {P_i \cos \left( {\frac{n}\pi } \right)} + \frac{2}{n}\sum\limits_{i = 0}^{{\tfrac{n}{2}} - 1} {Q_i \sin \left( {\frac{2i+1}{n}\pi } \right)}, & n:\text{even integer} \\ \\ \frac{1}{n}\ln \left( {1 + v} \right) - \frac{2}{n}\sum\limits_{i = 0}^ {P_i \cos \left( {\frac{2i+1}{n}\pi } \right)} + \frac{2}{n}\sum\limits_{i = 0}^ {Q_i \sin \left( {\frac{2i+1}{n}\pi } \right)}, & n:\text{odd integer} \\ \end{array} \right. $$ where

\begin{align} P_i &= \frac{1}{2}\ln \left( {v^2 - 2v\cos \left( {\frac{n}\pi } \right) + 1} \right) \\ Q_i &= \arctan \left( {\frac} \right) \end{align} $$