Given $\frac{dS}{dt}=-\frac{SI}{S + I + R}$ and $R = -(SI + S + I)/ \frac{dS}{dt}$ Determine $\frac{d^2S}{dt^2}$ I am given a system of equations:
\begin{align}
\frac{dS}{dt} &= -\frac{SI}{S + I + R}\\
\frac{dI}{dt} &= \frac{SI}{S + I + R} -I \\
\frac{dR}{dt} &= I - R\\
\end{align}
I am to subsitute $R = -\left(SI + S + I\right) / \frac{dS}{dt} $ to determine $\frac{d^2S}{dt^2}$
I chose to work only with the first equation.
When I substitute what is given for $R$ into the denominator I get:
\begin{align}
S + I + R &= S + I + R = -\left(SI + S + I\right) / \frac{dS}{dt}\\
&=\frac{\frac{dS}{dt}(S+I)-(SI+S+I)}{\frac{ds}{dt}}
\end{align}
I substitute this into the equation for $\frac{dS}{dt}$ and I get:
\begin{align}
\frac{dS}{dt} &= \frac{-SI \left(\frac{dS}{dt}\right)}{\frac{ds}{dt}\left(S+I\right)-(SI+S+I)}\\
\left(\frac{dS}{dt}\right)^2(S+I)-\frac{dS}{dt}\left(SI+S+I\right) &= -SI\left(\frac{dS}{dt}\right)\\
\left(\frac{dS}{dt}\right)^2(S+I) &= \left(\frac{dS}{dt}\right)\left(S+I\right)\\
\left(\frac{dS}{dt}\right)^2 &= \left(\frac{dS}{dt}\right)\\
\end{align}
How do I get the desired result of
$$
\frac{d^2S}{dt^2}= \frac{2}{S}\left(\frac{dS}{dt}\right)^2
$$
 A: Try differentiating $\frac{dS}{dt}$ directly.
$\begin{align}\frac{d^2S}{dt^2} &= -\frac{(S+I+R)\frac{d(SI)}{dt}- SI\frac{d(S+I+R)}{dt}}{(S+I+R)^2}\\
\\&=-\frac{(S+I+R)\left[I\frac{d(S)}{dt}+S\frac{d(I)}{dt}\right]- SI\frac{d(S+I+R)}{dt}}{(S+I+R)^2}\\
\\&=-\frac{(S+I+R)\left[-I\frac{SI}{S + I + R}+S\frac{SI}{S + I + R} -SI\right]+SIR}{(S+I+R)^2}\\
\\&=-\frac{(S+I+R)\left[\frac{-SI(R+2I)}{S + I + R}\right]+SIR}{(S+I+R)^2}\\
\\&=\frac{2SI^2}{(S+I+R)^2} \\
\\& = \frac{2}{S}\left(\frac{SI}{S+I+R}\right)^2\\
\\&=\frac{2}{S}\left(\frac{dS}{dt}\right)^2\end{align}$
A: I acknowledge this answer is very similar to that of our colleague Ak, with perhaps some differences in formatting and added words of explanation.
We start with the given equations
$\dot S = -\dfrac{SI}{S + I + R}, \tag 1$
$\dot I = \dfrac{SI}{S + I + R} - I, \tag 2$
$\dot R = I - R; \tag 3$
we differentiate (1) (here I have for the moment used "${}^\prime$" in lieu of "$\dot {}$" on the right-hand side):
$\ddot S = -\dfrac{(SI)'(S + I + R) - (SI)(S + I + R)'}{(S + I + R)^2}, \tag 4$
whence
$\ddot S = -\dfrac{(\dot S I + S\dot I)(S + I + R) - (SI)(\dot S + \dot I + \dot R)}{(S + I + R)^2}; \tag 5$
we add (1) and (2):
$\dot S + \dot I = -I, \tag 6$
and to this add (3):
$\dot S + \dot I + \dot R = -R, \tag 7$
and substitute this into (5):
$\ddot S = -\dfrac{(\dot S I + S\dot I)(S + I + R) + RSI}{(S + I + R)^2}; \tag 8$
again from (1) and (2):
$\dot SI + S\dot I = -\dfrac{SI^2}{S + I + R} + \dfrac{S^2I}{S + I + R} - IS$
$= \dfrac{S^2I - I^2S}{S + I + R} - IS, \tag 9$
whence
$(\dot SI + S\dot I)(S + I + R) = S^2I - I^2S - IS(S + I + R)$
$= S^2I - I^2S - IS^2 - I^2S - ISR = -2I^2S - ISR; \tag{10}$
thus,
$(\dot SI + S\dot I)(S + I + R) + RSI = -2I^2S; \tag{11}$
finally we substitute this into (8) and execute some very simple algebraic maneuvers:
$\ddot S = \dfrac{2I^2S}{(S + I + R)^2} = \dfrac{2I^2S^2}{S(S + I + R)^2}, \tag{12}$
$\ddot S = \dfrac{2}{S} \left (\dfrac{IS}{(S + I + R)} \right )^2 = \dfrac{2}{S} \dot S^2, \tag{13}$
$OE\Delta$.
