How to solve the second order ODE with Neumann boundary conditions?

Question

The ODE is defined on $[0,b]$ with Neumann boundary conditons. $$y''(x)=\frac{c_1y(x)}{c_2+y(x)}$$ $$y'(0)=0; y'(b)=0$$

How to solve the above ODE? Any help is appreciated!

Answer 1

Reduce the order of the equation by substituting $w = w(y)$. Then it yields

$$ww' = \frac{c_1y}{c_2 + y},$$

which is a separable ODE. This equation can be integrated in order to get

$$\frac{w^2}{2} = c_1 (y - c_2 \ln \lvert y + c_2 \rvert ) + C.$$

Solve this for $w$, then use

$$\int \frac{\mathrm dy}{w(y)} = \int dx$$

for calculating $y = y(x)$ and apply the boundary conditions.

Answer 2

$$y''(x)=\frac{c_1y(x)}{c_2+y(x)}$$ Multiply both sides by $2y'$: $$2y'y''(x)=\frac{2c_1y(x)y'}{c_2+y(x)}$$ $$(y'^2)'=\frac{2c_1y(x)y'}{c_2+y(x)}$$ Integrate: $$y'^2+K= 2c_1\int \frac{y(x)}{c_2+y(x)}dy$$ $$y'^2+K= 2c_1y-2c_1c_2\int \frac{dy}{c_2+y(x)}$$ $$y'^2+K= 2c_1y-2c_1c_2 \ln |{c_2+y(x)}|$$ Unfortunately, it's not easy to integrate. $$y'^2= 2c_1 \left (y-c_2 \ln |{c_2+y(x)}| \right )+C$$