Substituting a variable in the ODE: $ky''+cy'+y(1-y)=0$

Question

Let $k,c$ be constants. I would like to use a new variable $z=y'$ to make the following ODE into a first order ODE :

$ky''+cy'+y(1-y)=0$

I don't see how to handle the $y(1-y)$ term...

Any help is appreciated.

Answer 1

Contrarily to what one could say at first sight, Maple yields the answer in the form requested by the OP. One has to set $z=y(x)$ and this gives the equation $$ w(z)\frac{dw(z)}{dz}+\frac{1}{k}\left(c\cdot w(z)-z^2+z\right)=0 $$ and $$ x=\int\frac{dz}{w(z)}+C, $$ where $C$ is an integration constant and finally $$ w(z)=\frac{dy}{dx}. $$

Answer 2

You can transform it to a first order system by letting $z_1(x)=y(x)$ and $z_2(x) = y'(x)$. Then you will get with $z(x)=\begin{bmatrix}z_1(x)\\z_2(x)\end{bmatrix}$:

\begin{align*} z'(x) &= \begin{bmatrix}z_1'(x)\\z_2'(x)\end{bmatrix} = \begin{bmatrix}y'(x)\\y''(x)\end{bmatrix} = \begin{bmatrix}y'(x)\\-\frac{c}k y'(x) - \frac1k y(x)\bigl(1-y(x)\bigr)\end{bmatrix} \\ &= \begin{bmatrix}z_2(x)\\-\frac{c}k z_2(x) - \frac1k z_1(x)\bigl(1-z_1(x)\bigr)\end{bmatrix} \end{align*}

Answer 3

$$k\frac{d^2y}{dx^2}+c\frac{dy}{dx}+y(1-y)=0 \tag 1$$

This equation is on autonomous kind. The usual change in the aim to reduce the order from 2 to 1 is : $$\frac{dy}{dx}=F(y)$$ $\frac{d^2y}{dx^2}=\frac{dF(y)}{dx}=\frac{dF}{dy}\frac{dy}{dx}=F(y)\frac{dF}{dy}$ $$kF(y)\frac{dF}{dy}+cF(y)+y(1-y)=0$$ $$F(y)\frac{dF}{dy}=-\frac{c}{k}F(y)+\frac{1}{k}y(y-1)$$ Let $F(y)=\frac{1}{G(y)}$ $$\frac{dG}{dy}=-\frac{c}{k}G^2+\frac{1}{k}y(1-y)G^3$$ This is an Abel's differential equation of the first kind.

In the general case there is no closed form for the solutions. In practice, better use numerical method to directly solve Eq.$(1)$.

Note that obvious particular solutions of Eq.$(1)$ are $y(x)=0$ and $y(x)=1$. Of course, this is not the general solution.