Solution to y'=1-y^n

[XinZhong]

I have an ODE that I have worked on for a while, just cannot find a solution.


y(x)' = 1 - y(x)^n

where n is a real number larger than 1. The initial condition is x=0, y=0.

I tried hypergeometric function to describe x(y) but I specifically need y as a function of x.

Anybody has any idea?

PS The numerical solution looks like this with Euler method for n=3.



Many thanks.

Xin

 

[HallsofIvy]

This is a separable differential equation: \(\displaystyle \frac{dy}{dx}= 1- y^n\) can be written as \(\displaystyle \frac{dy}{1- y^n}= dx\). If your problem is integrating the left side, the denominator can be factored as \(\displaystyle (1- y)(1- y+ y^2- y^3+ \cdot\cdot\cdot+ (-1)^{n-1}y^{n-1}\)

 

[XinZhong]

This is a separable differential equation: \(\displaystyle \frac{dy}{dx}= 1- y^n\) can be written as \(\displaystyle \frac{dy}{1- y^n}= dx\). If your problem is integrating the left side, the denominator can be factored as \(\displaystyle (1- y)(1- y+ y^2- y^3+ \cdot\cdot\cdot+ (-1)^{n-1}y^{n-1}\)

Thank you for your reply. The problem of doing so is 1) n is not necessarily an integer, and 2) the integration can be done but is it possible to write y as a function as x?
Sorry if I may not have made it clear enough. I was looking for an expression y as a function of x that satisfies \(\displaystyle \frac{dy}{dx}= 1- y^n\)