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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?

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  • $\begingroup$ Provide some boundary conditions and no one will tell you y = 0 is a solution :). Boundary conditions will also influence the solution process (if it is possible). Also, if the problem is an initial value one, do you need the solution for all x or will large x do? $\endgroup$
    – Paul
    Commented Apr 4, 2017 at 11:07
  • $\begingroup$ One initial condition is y'(0)=0. (This is a second order ODE related to a Riccati equation with initial condition u(0)=0). I need the solution for all x actually. I see that for large x the exponential term fades away. Thank you :) $\endgroup$
    – NSZ
    Commented Apr 4, 2017 at 11:12

1 Answer 1

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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

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  • $\begingroup$ Thank you for your answer. However I didn't really understand your first subsitution because I would get only $y''(x)+(a_1+a_2z)y'(x)+(b_1+b_2z+b_3z^2)y=0$. And also what is the function $g(\xi)$? $\endgroup$
    – NSZ
    Commented Apr 4, 2017 at 12:48
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    $\begingroup$ you also need to transform the derrivatives, like $\frac{d}{dx}=\frac{dz}{dx}\frac{d}{dz}=-\gamma{z}\frac{d}{dz}$, etc! $g(\xi)$ is the function satisfying the confluent hypergeometric equation. $\endgroup$ Commented Apr 4, 2017 at 13:57
  • $\begingroup$ @Kyril Pesotski Do you maybe mean the substitution $y=z^{-\frac{-\gamma a_1+\gamma^2}{2\gamma^2}}e^{-\frac{a_2}{\gamma^2}}g(\xi)$, with z instead of x? $\endgroup$
    – NSZ
    Commented Apr 11, 2017 at 8:37

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