Solution to Nonlinear ODE $s''( t) s( t) =( s'( t))^{2} +B( s( t))^{2} s'( t) -g\cdot s( t) s'( t)$

I've solved linear ODEs before. This however is something completely new to me. I want to solve it without using approximations or anything.

$s''( t) s( t) =( s'( t))^{2} +B( s( t))^{2} s'( t) -g\cdot s( t) s'( t)$

These are the equations I started with

$ \begin{array}{l} s'( t) =-Bs( t) i( t)\\ r'( t) =g\cdot i( t)\\ i'( t) =i( t)( Bs( t) -g) \end{array}$

The reason I'm solving this particular equation is because I want to solve for all of the functions $(i(t),r(t),s(t))$ from above.

I just figured $s(t)$ might be the place to start. I really don't have any clue where to start here. Any help would be awesome! Thanks!

--edit-- I made a mistake above. In order to fix it I changed $i'( t) =i( t)( Bs( t) -1)$ to this $i'( t) =i( t)( Bs( t) -g)$. My bad...

---Edit---

I might have a step towards the answer.

$ \begin{array}{l} v( s) =s'( t)\\ \Longrightarrow \\ s''( t) =\frac{dv( s)}{dt} =\frac{dv}{ds}\frac{ds}{dt} =v'( s) \cdot \frac{ds}{dt} =v'( s) \cdot v( s)\\ \Longrightarrow \\ v'( s) \cdot v( s) s=v( t)^{2} +Bs^{2} v( t) -g\cdot s\cdot v( t)\\ \Longrightarrow \\ v'( s) \cdot s=v( t) +Bs^{2} -g\cdot s \end{array}$

Wolfram Alpha says that the solution to this differential equation is

$v(s)=Bs^2+c_1s-gsln(s)$

So that means that

$s'(t)=Bs(t)^2+c_1s(t)-gs(t)\ln(s(t))$

This is certainly better than before

Me and my friend got it! (This was also exactly what Sal in the comments suggested doing) $ \begin{array}{l} v( s) =s'( t) \Longrightarrow \\ \frac{dv( s)}{dt} =\frac{dv( s)}{ds} \cdot \frac{ds}{dt} =v'( s) \cdot v( s) \ \text{literally the chain rule}\\ \Longrightarrow \\ v'( s) \cdot v( s) \cdot s=v( s)^{2} +Bs^{2} v( s) -gv( s) s\\ \\ \therefore v'( s) s=v( s) +Bs^{2} -gs \end{array}$

Next find an integrating factor.

$ \begin{array}{l} v'( s) -\frac{v( s)}{s} =Bs-g\\ integrating\ factor\ I=e^{\int \frac{1}{s} ds} =\frac{1}{s} \end{array}$

Multiply both sides of the equation by the integrating factor.

$Iv'( s) -I\frac{1}{s} v( s) =I( Bs-g)$

The really interesting thing that makes this work is $\frac{d} {ds}(Iv(s))=Iv'(s)-I\frac{1}{s}v(s)$

$ \begin{array}{l} \therefore \frac{d( Iv( s))}{dt} =I( Bs-g)\\ \Longrightarrow \\ Iv( s) =\int I( Bs-g) ds=\int \left( B-\frac{g}{s}\right) ds=Bs-ln( s) +c_{1} \end{array}$

So we get

$v( s) =Bs^{2} +c_{1} s-gsln( s) \Longrightarrow s'( t) =Bs( t)^{2} +c_{1} s( t) -gs( t) ln( s( t))$

Next just split the derivative and integrate

$1dt=\frac{1}{Bs( t)^{2} +c_{1} s( t) -gs( t) ln( s( t))} ds \Longrightarrow t+c_{2} =\int \frac{1}{Bs( t)^{2} +c_{1} s( t) -gs( t) ln( s( t))} ds$

Solving for the constants we get

$ \begin{array}{l} at\ t=0\\ s'( 0) =Bs( 0)^{2} +c_{1} s( 0) -gs( 0) ln( s( 0))\\ and\ we\ know\ that\ s'( t) =-Bs( t) i( t)\\ \\ s'( 0) =-Bs( 0) i( 0)\\ \Longrightarrow \\ -Bs_{0} i_{0} =Bs_{0}^{2} +c_{1} s_{0} -gs_{0} ln( s_{0})\\ \Longrightarrow \\ c_{1} =gln( s_{0}) +B( s_{0} -i_{0})\\ c_{2} =\int _{0}^{s( 0)}\frac{1}{B\omega ^{2} +c_{1} \omega -g\omega ln( \omega )} d\omega =\int _{0}^{s_{0}}\frac{1}{B\omega ^{2} +( gln( s_{0}) +B( s_{0} -i_{0})) \omega -g\omega ln( \omega )} d\omega \\ \\ \therefore t=\\ \int\limits _{0}^{s( t)}\frac{1}{B\omega ^{2} +( gln( s_{0}) +B( s_{0} -i_{0})) \omega -g\omega ln( \omega )} d\omega -\int\limits _{0}^{s_{0}}\frac{1}{B\omega ^{2} +( gln( s_{0}) +B( s_{0} -i_{0})) \omega -g\omega ln( \omega )} d\omega \end{array}$

We can combine the integrals because of their limits

$t=\int _{s_{0}}^{s( t)}\frac{1}{B\omega ^{2} +( gln( s_{0}) +B[ s_{0} -i_{0}]) \omega -g\omega ln( \omega )} d\omega $

Desmos graphs this in its current form, but if you wanted $s(t)=$something then the Lagrange Inversion theorem probably works here.

Edit... I was actually able to graph all three functions for the SIR model in desmos (green equation is s(t), red i(t) blue r(t)) desmos graph for SIR model