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

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  • $\begingroup$ 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. $\endgroup$
    – Christoph
    Commented Jan 16, 2019 at 2:31

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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.

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