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Solve this Differential Equation: u"(x) = (lambda * (e^{-x} / x - k^2) u(x) = 0
- Thread starter mario99
- Start date
Please share some of the trials that you have conducted using your skills'[math]u''(x) + \left(\lambda\frac{e^{-x}}{x}- k^2\right)u(x) = 0[/math]
I used all my skills and every trick I know in solving differential equations, but I failed to solve the above differential equation.
Let [imath]\displaystyle \ t = \frac{e^{-x}}{x} \ \text{and} \ u(x) = y(t(x))[/imath]
[imath]\displaystyle \frac{du}{dx} = -\frac{dy}{dt}\frac{(x+1)e^{-x}}{x^2}[/imath]
[imath]\displaystyle \frac{d^2u}{dx^2} = -\frac{dy}{dt}\frac{(x^2+2x+2)e^{-x}}{x^3} - \frac{d^2y}{dt^2}\left(\frac{(x+1)e^{-x}}{x^2}\right)^2[/imath]
[imath]\displaystyle -\frac{dy}{dt}\frac{(x^2+2x+2)e^{-x}}{x^3} - \frac{d^2y}{dt^2}\left(\frac{(x+1)e^{-x}}{x^2}\right)^2 - \left(\lambda\frac{e^{-x}}{x}- k^2\right)\frac{dy}{dt}\frac{(x+1)e^{-x}}{x^2} = 0[/imath]
[imath]\displaystyle -\frac{dy}{dt}\frac{(x^2+2x+2)t}{x^2} - \frac{d^2y}{dt^2}\left(\frac{(x+1)t}{x}\right)^2 - \left(\lambda t- k^2\right)\frac{dy}{dt}\frac{(x+1)t}{x} = 0[/imath]
[imath]\displaystyle \frac{d^2y}{dt^2}\left(\frac{(x+1)t}{x}\right)^2 + \frac{dy}{dt}\frac{(x^2+2x+2)t}{x^2} + \left(\lambda t- k^2\right)\frac{dy}{dt}\frac{(x+1)t}{x} = 0[/imath]
[imath]\displaystyle t^2\frac{d^2y}{dt^2} + t\frac{dy}{dt}\left(1 + \frac{1}{(x + 1)^2}\right) + t\frac{dy}{dt}\left(\frac{\lambda xt- xk^2}{x+1}\right) = 0[/imath]
[imath]\displaystyle t^2\frac{d^2y}{dt^2} + t\frac{dy}{dt}\left[\left(1 + \frac{1}{(x + 1)^2}\right) + \left(\frac{\lambda xt- xk^2}{x+1}\right)\right] = 0[/imath]
[imath]\displaystyle \frac{du}{dx} = -\frac{dy}{dt}\frac{(x+1)e^{-x}}{x^2}[/imath]
[imath]\displaystyle \frac{d^2u}{dx^2} = -\frac{dy}{dt}\frac{(x^2+2x+2)e^{-x}}{x^3} - \frac{d^2y}{dt^2}\left(\frac{(x+1)e^{-x}}{x^2}\right)^2[/imath]
[imath]\displaystyle -\frac{dy}{dt}\frac{(x^2+2x+2)e^{-x}}{x^3} - \frac{d^2y}{dt^2}\left(\frac{(x+1)e^{-x}}{x^2}\right)^2 - \left(\lambda\frac{e^{-x}}{x}- k^2\right)\frac{dy}{dt}\frac{(x+1)e^{-x}}{x^2} = 0[/imath]
[imath]\displaystyle -\frac{dy}{dt}\frac{(x^2+2x+2)t}{x^2} - \frac{d^2y}{dt^2}\left(\frac{(x+1)t}{x}\right)^2 - \left(\lambda t- k^2\right)\frac{dy}{dt}\frac{(x+1)t}{x} = 0[/imath]
[imath]\displaystyle \frac{d^2y}{dt^2}\left(\frac{(x+1)t}{x}\right)^2 + \frac{dy}{dt}\frac{(x^2+2x+2)t}{x^2} + \left(\lambda t- k^2\right)\frac{dy}{dt}\frac{(x+1)t}{x} = 0[/imath]
[imath]\displaystyle t^2\frac{d^2y}{dt^2} + t\frac{dy}{dt}\left(1 + \frac{1}{(x + 1)^2}\right) + t\frac{dy}{dt}\left(\frac{\lambda xt- xk^2}{x+1}\right) = 0[/imath]
[imath]\displaystyle t^2\frac{d^2y}{dt^2} + t\frac{dy}{dt}\left[\left(1 + \frac{1}{(x + 1)^2}\right) + \left(\frac{\lambda xt- xk^2}{x+1}\right)\right] = 0[/imath]
I hate to use my last resource (Power Series) to solve this problem because it is very long, complicated, and most of the time it gives no clear pattern to recognize the functions it converges to.
[imath]\displaystyle u = \sum_{n=0}^{\infty}a_nx^n[/imath]
[imath]\displaystyle u' = \sum_{n=0}^{\infty}na_nx^{n-1}[/imath]
[imath]\displaystyle u'' = \sum_{n=0}^{\infty}n(n-1)a_nx^{n-2}[/imath]
---------------------------------------
[imath]\displaystyle u''(x) + \left(\lambda\frac{e^{-x}}{x}- k^2\right)u(x) = \sum_{n=0}^{\infty}n(n-1)a_nx^{n-2} + \left(\lambda\frac{e^{-x}}{x}- k^2\right)\sum_{n=0}^{\infty}a_nx^n = 0[/imath]
[imath]\displaystyle \sum_{n=0}^{\infty}n(n-1)a_nx^{n-2} + \sum_{n=0}^{\infty}\lambda e^{-x}a_nx^{n-1} - \sum_{n=0}^{\infty}k^2a_nx^n = 0[/imath]
How to combine the three summations into one summation??
[imath]\displaystyle u = \sum_{n=0}^{\infty}a_nx^n[/imath]
[imath]\displaystyle u' = \sum_{n=0}^{\infty}na_nx^{n-1}[/imath]
[imath]\displaystyle u'' = \sum_{n=0}^{\infty}n(n-1)a_nx^{n-2}[/imath]
---------------------------------------
[imath]\displaystyle u''(x) + \left(\lambda\frac{e^{-x}}{x}- k^2\right)u(x) = \sum_{n=0}^{\infty}n(n-1)a_nx^{n-2} + \left(\lambda\frac{e^{-x}}{x}- k^2\right)\sum_{n=0}^{\infty}a_nx^n = 0[/imath]
[imath]\displaystyle \sum_{n=0}^{\infty}n(n-1)a_nx^{n-2} + \sum_{n=0}^{\infty}\lambda e^{-x}a_nx^{n-1} - \sum_{n=0}^{\infty}k^2a_nx^n = 0[/imath]
How to combine the three summations into one summation??
The similar way you did it in Bessel's function or Airy's stress function!How to combine the three summations into one summation?
The Bessel function, I know it, but the Airy stress function, I have never heard of it before. And if you meant the Bessel and Airy equations, that's another thing.The similar way you did it in Bessel's function or Airy's stress function!
Let us assume that you meant the latter. Yeah I have solved the Airy equation from scratch successfully and tried to solve the Bessel equation as well, but had some issues. Solving them was not similar to the differential equation above in post #1.
Exactly in which term of the Bessel or the Airy equation, did you see the exponential function [imath]e^{-x}[/imath]??