I'm trying to understand a proof from this paper on the SIR model. This appears at the bottom of page four.
The following two equations are given:
$ u\frac{d^2u}{dt^2} - (\frac{du}{dt})^2 + (\gamma - x_0\beta u)u\frac{du}{dt} = 0 \quad \mathbf{(22)}$
$ \phi = \frac{dt}{du} \qquad\qquad\qquad\qquad\qquad\qquad\quad\; \mathbf{(23)}$
The next part is where I am having trouble. It reads:
With the help of the transformation given by Eq. (23), Eq. (22) becomes a Bernoulli type differential equation,
$$ \frac{d\phi}{du} + \frac{1}{u}\phi = (\gamma - x_0\beta u)\phi^2 \quad \mathbf{(24)}$$
How is this possible? I can't seem to work it out. Please show me how equations (22) and (24) are equivalent.
Because you have: $$\frac {du}{dt}=\frac 1 {\phi}$$ Therefore: $$\frac {d^2u}{dt^2}=\frac {d}{dt}\frac 1 {\phi}= \left (\frac {d}{du}\frac 1 {\phi} \right )\frac {du}{dt}$$ $$\frac {d^2u}{dt^2}=\frac 1 {\phi}\left (\frac {d}{d \phi}\frac 1 {\phi} \right)\frac {d\phi}{du}$$ $$\frac {d^2u}{dt^2}=\frac 1 {\phi} \left(-\frac 1 {\phi ^2} \right)\frac {d\phi}{du}$$ $$\frac {d^2u}{dt^2}=-\frac 1 {\phi^3}\frac {d\phi}{du} \text { and } \frac {du}{dt}=\frac 1 {\phi}$$
Substitute that in your original equation : $$u\frac{d^2u}{dt^2} - (\frac{du}{dt})^2 + (\gamma - x_0\beta u)u\frac{du}{dt} = 0 $$ The equation becomes: $$\frac{d\phi}{du} + \frac{1}{u}\phi - (\gamma - x_0\beta u)\phi^2=0 $$