I need to solve the following differential equation:
$$y''(x)+\left(a_1+a_2e^{-\gamma x}\right) y'(x)+ \left(b_1+b_2e^{-\gamma x}+b_3 e^{-2\gamma x}\right) y(x)=0 $$
It is linear, the expression is plain and clean, but the coefficient are not constant, which constitutes a big problem.
What is the approach I have to use? Does someone know how to solve it?
Unfortunately I didn't find anything in the literature but I hope this community will surprise me.
EDIT: From this point on I continued thanks to Kiryl Pesotski's suggestion.
If we apply the substitution $z=e^{-\gamma x}$, we have that: $$\frac{d}{dx}y(x)=\frac{dz}{dx}\frac{d}{dz}y(-\log(z)/\gamma)=-\gamma e^{-\gamma x}\frac{d}{dz}y(-\log(z)/\gamma)=-\gamma z \frac{d}{dz}y(-\log(z)/\gamma)$$ where, for simplicity, we can rename $y(-\log(z)/\gamma)$ as $y(z)$. Then: $$\frac{d^2}{dx^2}y(x)=\frac{dz}{dx}\frac{d}{dz}\left(-\gamma z y'(z)\right)=\cdots =\gamma^2 z(y'(z)+zy''(z))$$ The ode becomes: $$z^2 y''(z)+\frac{1}{\gamma}zy'(z)\left(\gamma-a_1-a_2z\right)+\frac{1}{\gamma^2}y(z)(b_1+b_2z+b_3z^2)=0$$ Now I exploit the second suggestion because I am supposed to get an hypergeometric equation. The substitution $y(z)=x^{-\frac{\gamma-a_1}{2\gamma}}e^{-\frac{a_2}{\gamma^2}}g(\xi)$ with $\xi=\sqrt{\frac{b_1+\gamma^2}{\gamma^2}-\frac{b_3}{\gamma^2}}z$ has the following first derivative: $$\frac{d}{dz}y=\frac{1}{\gamma z}\frac{\gamma-a_1}{\gamma}x^{-\frac{\gamma-a_1}{2\gamma}-1}e^{-\frac{a_2}{\gamma^2}}g(\xi)+x^{-\frac{\gamma-a_1}{2\gamma}}e^{-\frac{a_2}{\gamma^2}}g'(\xi)\sqrt{\frac{b_1+\gamma^2}{\gamma^2}-\frac{b_3}{\gamma^2}}$$ However, if I have not made any mistake, I don't understand how from virtue of this substitution I can get to the final solution.
Can someone know how to complete the passages and get the hypergeometric function solution of this equation?
The solution is given in terms of the Confluent Hypergeometric functions! First you substitute $$z=e^{-\gamma{x}}$$ The equation becomes $$z^{2}y''+(\frac{-\gamma{a_{1}}+\gamma^{2}}{\gamma^{2}}-({a_{2}}/\gamma){z})zy'+(\frac{b_{1}}{\gamma^{2}}+\frac{b_{2}}{\gamma^{2}}z+\frac{b_{3}}{\gamma^{2}}z^{2})y=0$$ Now by virtue of substitution $y=x^{-\frac{-\gamma{a}_{1}+\gamma^{2}}{2\gamma^{2}}}e^{-a_{2}/\gamma2}g(\xi)$ with $\xi=\sqrt{\frac{b_{1}+\gamma^{2}}{\gamma^{2}}-4\frac{b_{3}}{\gamma^{2}}}z$ you arrive to the confluent hypergeometric equation