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 ${\textstyle (y')^{m}=(x+y)^{k}}$. Substituting $v=x+y$ and its derivative $v'=1+y'$ gives ${\textstyle (v'-1)^{m}=v^{k}}$, which can be written ${\textstyle {\tfrac {dv}{dx}}=1+v^{\tfrac {k}{m}}}$, 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 ${\textstyle \int {\frac {dv}{1+v^{n}}}=x+C}$. In the tables book Gradshteyn and Ryzhik, this form decomposes as:

$\int {\frac {dv}{1+v^{n}}}=\left\{{\begin{array}{ll}-{\frac {2}{n}}\sum \limits _{i=0}^{{\textstyle {n \over 2}}-1}{P_{i}\cos \left({{\frac {2i+1}{n}}\pi }\right)}+{\frac {2}{n}}\sum \limits _{i=0}^{{\tfrac {n}{2}}-1}{Q_{i}\sin \left({{\frac {2i+1}{n}}\pi }\right)},&n:{\text{even integer}}\\\\{\frac {1}{n}}\ln \left({1+v}\right)-{\frac {2}{n}}\sum \limits _{i=0}^{\textstyle {{n-3} \over 2}}{P_{i}\cos \left({{\frac {2i+1}{n}}\pi }\right)}+{\frac {2}{n}}\sum \limits _{i=0}^{\tfrac {n-3}{2}}{Q_{i}\sin \left({{\frac {2i+1}{n}}\pi }\right)},&n:{\text{odd integer}}\\\end{array}}\right.$

where

${\begin{aligned}P_{i}&={\frac {1}{2}}\ln \left({v^{2}-2v\cos \left({{\frac {2i+1}{n}}\pi }\right)+1}\right)\\Q_{i}&=\arctan \left({\frac {v-\cos \left({{\textstyle {{2i+1} \over n}}\pi }\right)}{\sin \left({{\textstyle {{2i+1} \over n}}\pi }\right)}}\right)\end{aligned}}$

See also

• Examples of differential equations