Following the work of Yaoji Lu - 1967 (here's a link to the full paper) I got stuck at the step when the author transform a non-linear differential equation into a linear equation (eq. 3.9 pag. 19).
More precisely he starts from this equation:
$$\frac{dY}{dX}=\frac{Y}{X}-\alpha \cdot X^{-\frac{c}{b}-1}\cdot Y^{\frac{1}{b}}$$
Then define $Z=Y^{1-\frac{1}{b}}$ and eventually get:
$$\frac{dZ}{dX}+\frac{1-b}{b}\cdot \frac{1}{X}\cdot Z=\alpha \frac{1-b}{b}\cdot X^{-\frac{c}{b}-1}$$
Is there anyone who can help me with this?
It's Bernouilli's equation $$\frac{dY}{dX}=\frac{Y}{X}-\alpha \cdot X^{-\frac{c}{b}-1}\cdot Y^{\frac{1}{b}}$$ Divide by $y^{1/b}$ $$Y^{-\frac{1}{b}}Y'=\frac{Y^{1-\frac 1b}}{X}-\alpha \cdot X^{-\frac{c}{b}-1} $$ $$(1-\frac 1b )Y^{-\frac{1}{b}}Y'=(1-\frac 1b)(\frac{Y^{1-\frac 1b}}{X}-\alpha \cdot X^{-\frac{c}{b}-1} )$$ $$(Y^{1-\frac{1}{b}})'=(1-\frac 1b)(\frac{Y^{1-\frac 1b}}{X}-\alpha \cdot X^{-\frac{c}{b}-1} )$$ Substitute $Z=Y^{1-\frac 1b}$ $$Z'=(1-\frac 1b)(\frac Z{X}-\alpha \cdot X^{-\frac{c}{b}-1} )$$