# Solution to a second-order nonlinear ordinary differential equation

I am given three functions $$z'(t), y'(t), x'(t)$$ and I have reduced these functions to the equation: $$$$z''=x_{0}\beta z'e^{-\frac{\beta}{\gamma}z} - \gamma z$$$$ where $$\beta, \gamma ,x_{0}$$ are constants. Next we introduce a new function: $$$$u(t)=e^{-\frac{\beta}{\gamma}z}$$$$ Substitution yields $$$$u\frac{d^{2}u}{dt^{2}}- \bigg(\frac{du}{dt}\bigg)^{2}+(\gamma - x_{0}\beta u)u\frac{du}{dt} = 0$$$$ Next a new function is introduced: $$$$\phi = \frac{dt}{du}$$$$ Now the equation can be rearranged to: $$$$\frac{d\phi}{du}+\frac{1}{u} \phi = (\gamma-x_{0}\beta u )\phi^{2}$$$$ The solution to this equation is: $$$$\phi = \frac{1}{u(C_{1}-\gamma \ln u +x_{0}\beta u)}$$$$ So what would be the solution to the original equation in terms of $$u$$?

• The ordinary differential equation (ODE) that you write down for $u$ is equivalent to $z'' = x_0 \beta z' e^{-\frac{\beta}{\gamma} z} - \gamma z'$. So there is either a mistake in the original ODE for $z$ or in the ODE for $u$. Please check. Commented Jan 16, 2019 at 2:31

The last equation od $$\frac{dt}{du}=\frac{1}{u(C_{1}-\gamma \ln u +x_{0}\beta\, u)}.$$ Integrating $$t=\int\frac{du}{u(C_{1}-\gamma \ln u +x_{0}\beta\, u)}.$$ I do not think that there is a closed formula for the integral.