I am a master student in mathematical physics. I study soliton and traveling wave solutions of the differential equations.
Let's consider the following ODE: $$Q^{\prime}(\xi)=ln(A)(\alpha+\beta Q(\xi)+\sigma Q^2(\xi))$$ where $A \neq 0,1.$
In a book, the solutions of the ODE are given as follows: (But, I don't understand how to derive it.)
There are twelve solution cases w.r.t coefficients of ODE. Any help would be appreciated.
CASE I) When $\beta^{2}-4 \alpha \sigma<0$ and $\sigma \neq 0$, then $$ \begin{array}{l} Q_{1}(\xi)=-\frac{\beta}{2 \sigma}+\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{2 \sigma} \tan _{A}\left(\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{2} \xi\right) \\ Q_{2}(\xi)=-\frac{\beta}{2 \sigma}-\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{2 \sigma} \cot _{A}\left(\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{2} \xi\right) \\ Q_{3}(\xi)=-\frac{\beta}{2 \sigma}+\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{2 \sigma}\left(\tan _{A}\left(\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)} \xi\right) \pm \sqrt{p q} \sec _{A}\left(\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right) \xi}\right)\right) \\ Q_{4}(\xi)=-\frac{\beta}{2 \sigma}-\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{2 \sigma}\left(\cot _{A}\left(\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)} \xi\right) \pm \sqrt{p q} \csc _{A}\left(\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)} \xi\right)\right) \\ Q_{5}(\xi)=-\frac{\beta}{2 \sigma}+\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{4 \sigma}\left(\tan _{A}\left(\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{4} \xi\right)-\cot _{A}\left(\frac{\sqrt{-\left(\beta^{2}-4 \alpha\right) \sigma}}{4} \xi\right)\right) \end{array} $$
CASE II:
$\vdots$
CASE XII: When $\beta=\lambda, \sigma=m \lambda(m \neq 0)$ and $\alpha=0,$ then $$ Q_{37}(\xi)=\frac{p A^{\lambda \xi}}{q-m p A^{\lambda \xi}} $$ where triangular functions are defined as \begin{array}{l} \sin _{A}(\xi)=\frac{p A^{i \xi}-q A^{-i \xi}}{2 i}, \quad \cos _{A}(\xi)=\frac{p A^{i \xi}+q A^{-i \xi}}{2} \\ \tan _{A}(\xi)=-i \frac{p A^{i \xi}-q A^{-i \xi}}{p A^{i \xi}+q A^{-i \xi}}, \quad \cot _{A}(\xi)=i \frac{p A^{i \xi}+q A^{-i \xi}}{p A^{i \xi}-q A^{-i \xi}} \\ \sec _{A}(\xi)=\frac{2}{p A^{i \xi}+q A^{-i \xi}}, \quad \csc _{A}(\xi)=\frac{2 i}{p A^{i \xi}-q A^{-i \xi}} \end{array} where $\xi$ is an independent variable, $p$ and $q$ are constants greater than zero and called deformation parameters.
$$Q^{\prime}(\xi)=ln(A)(\alpha+\beta Q(\xi)+\sigma Q^2(\xi))$$ In order to simplify the typing, change of symbols : $\begin{cases} Q=y \\ \xi=x \\ \ln(A)\alpha=a \\ \ln(A)\beta=b \\ \ln(A)\sigma=c \end{cases}$ $$y'(x)=a+by(x)+c\big(y(x)\big)^2$$ This is a Riccati ODE. The usual change of function to solve it is : $$y(x)=\frac{-1}{c}\frac{u'(x)}{u(x)}$$ which leads to : $$u''(x)-b\:u'(x)+a\:c\:u(x)=0$$ This is a second order linear ODE with constant coefficients. I suppose that you can continue.