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I have been given a second-order differential equation of the form: \begin{equation} \frac{d^2z}{dt^2}=x_0\beta \,\frac{dz}{dt}\,e^{-\beta z/\gamma} - \gamma \,\frac{dz}{dt} \end{equation} where, $ x_{0}, \beta, \gamma $ are constants. By introducing the function \begin{equation} u=e^{-\beta z/\gamma} \end{equation} and substituting it into the equation, how can one achieve the form: \begin{equation} u\frac{d^{2}u}{dt^{2}}-\bigg(\frac{du}{dt}\bigg)^{2}+\bigg(\gamma-x_{0}\beta u\bigg)u\frac{du}{dt} = 0? \end{equation}

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  • $\begingroup$ Does $'$ denote differentiation with respect to $t$? $\endgroup$ – John Doe Jan 9 at 14:24
  • $\begingroup$ Sorry for not clarifying, yes indeed everything is differentiated with respect to $t$ $\endgroup$ – nipohc88 Jan 9 at 14:27
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We have the equation $$z''=(x_0\beta u-\gamma)z'\tag1$$ Given that substitution, we can compute $z'$ and $z''$ in terms of derivatives of $u$. $$z=-\frac\gamma\beta\log u\\z'=-\frac\gamma{\beta u}u'\\z''=\frac\gamma{\beta u^2}(u')^2-\frac\gamma{\beta u}u''$$Substitute these results into $(1)$ and multiply through by $-\frac{\beta u^2}\gamma$ to obtain the desired equation.

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