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Is there any way to find an analytic solution to the following differential equation?

$$w'' - \left( \beta e^{-t} - 2 \right)w' + \gamma e^{-2t} w = 0$$

Here, $\gamma$ is negative and $\beta$ could be set to zero but preferably not.

I tried to solve it with $x = e^{-t}\sqrt{\gamma}$ to reduce it to a Bessel equation of order 1 before I realized that $\gamma$ was negative and hence the change of variable is not possible.

Thanks for your help.

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  • $\begingroup$ You can allow for it to be complex to aid in analysis. $\endgroup$ – shai horowitz May 16 '16 at 20:09
  • $\begingroup$ Cannot be complex, $\beta$ and $\gamma$ have an actual meaning in our model and represent real parameters. $\endgroup$ – user2059456 May 17 '16 at 3:17
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HINT :

enter image description here

If $\beta=0$ this is a Bessel ODE and the solution is straightforward.

If $\beta\neq 0$, this kind of second order linear ODE can be reduced to a confluent hypergeometric ODE thanks to a change of function such as : $$w(x)=x^ae^{bx}y(x)$$

This is the most arduous and tiresome part of the work.

After this change of function, the new ODE is turned on the form : $$xy''+(C-x)y'-Ay=0$$ in derermining the parameters $a$ and $b$ as functions of $\beta$ and $\gamma$ so that the form of the new ODE fit with the confluent hypergeometric equation :

http://mathworld.wolfram.com/ConfluentHypergeometricDifferentialEquation.html

The solutions, expressed as a linear combination of two confluent hypergeometric functions (or eventualy associated Legendre polynomial) allows to come backward to the solutions $w(t)$ of the initial ODE.

Sorry, since I am about to leave for several days, I have no more available time for more details.

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  • $\begingroup$ Thanks! Quick question though. When $\beta = 0$ first derivative is multiplied with $-x$, whereas bessel function has $+ x$ factor. Could you please specify if this is a modified Bessel function and what order it is. Thanks $\endgroup$ – user2059456 May 17 '16 at 19:49
  • $\begingroup$ Also here $\gamma$ is negative we can let $\gamma = - a^2$ So if $\beta = 0$, we have : $\frac{d^2w}{dx^2} - \frac{1}{x} \frac{dw}{dx} - a^2w =0$ $\endgroup$ – user2059456 May 17 '16 at 20:39
  • $\begingroup$ The general solution of $\quad w''-\frac{1}{x}w'-a^2w=0\quad$ is $\quad w=c_1x\,I_{1}(ax)+c_2x\,I_{-1}(ax)\quad$ where $I_{1}$ and $I_{-1}$ are the modified Bessel functions of order $1$ and $-1$ respectively. $\endgroup$ – JJacquelin May 24 '16 at 14:33

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