Solving mathematical differential equation

Question

I am struggling in solving this differential equation:

$$b'(x)=b(x)^{2}\: +\: b(x)\ln(b(x))(K-\gamma ).$$

Does anyone have any suggestions?

Answer 1

The presence of log y renders analytically solving it impossible, so a numerical solution is suggested with one differentiation ( primes with respect to $x$ ) to raise ODE order to two, eliminating constant $(K-\gamma).$

Also it is customary to use $(a,b,c)$ for constants. Let $ y(x)= b(x)$ and let primes denote differentiation with respect to $x$. Employ Chain Rule..

$$\dfrac{y'/y-y}{\log\,y}=\dfrac{\dfrac{{yy''-y^{'2}}}{y^2}-1}{\dfrac{1}{y}}$$

$$ RHS=y''-\dfrac{y^2+y^{'2}}{y}$$

so the ODE is:

$$ y''= y+y^{'2}/y + \dfrac{y'/y-y}{\log\,y}$$

can be integrated with convenient boundary conditions:

$$ x=0,y=1, y'=1. $$