# Modification of a differential equation

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: $$\begin{equation} u\frac{d^{2}u}{dt^{2}}-\bigg(\frac{du}{dt}\bigg)^{2}+\bigg(\gamma-x_{0}\beta u\bigg)u\frac{du}{dt} = 0? \end{equation}$$

• Does $'$ denote differentiation with respect to $t$? – John Doe Jan 9 at 14:24
• Sorry for not clarifying, yes indeed everything is differentiated with respect to $t$ – nipohc88 Jan 9 at 14:27

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.