Ordinary differential equation/solving with Frobenius method

[matematicar73]

Hello,
since two days I'm trying to get trough it but somehow always stuck at the same point during the process.

differential equation is: x^3*y" + x*y' + d*y = 0

which has an irregular singular point at x=0 (I know how to prove it), and even though it is an irregular point a solution is possible to obtain using Frobenius method of the Frobenius form: y=sum_(n=0)^∞ (a_n * x^(n+v))

Now what I've done so far:
y = sum_(n=0)^∞ [a_n * x^(n+v)]
y' = sum_(n=0)^∞ [a_n * (n+v) * x^(n+v-1)]
y" = sum_(n=0)^∞ [a_n * (n+v-1) * (n+v) * x^(n+v-2)]

I plug it into the [FONT=Helvetica Neue, Helvetica, Arial, sans-serif]differential equation and got so far:
sum_(n=0)^[/FONT]∞ [(n+v)*(n+v-1)*a_n*x^(n+v-2)] + sum_(n=0)^[FONT=Helvetica Neue, Helvetica, Arial, sans-serif]∞ [(n+v+d) * a_n * x^(n+v-3) = 0[/FONT]

[FONT=Helvetica Neue, Helvetica, Arial, sans-serif]So I got indicial equation: (v+d) * a_0 = 0 and since a_0 is not zero, v+d=0 which gives v= -d
and recursion equation: a_n = [ - (n-d-1) * (n-d-2) * a_(n-1) ] / n

Am I right so far? HELP!

The point is now that I have to put d = - 1 and I have to get the solution where a_n = (n-2) * a_(n-1) which I don't get

further more there are given boundary condition y(1)=0 and y(2)=1 and I have to show that the particular solution is y= (3 * e - 3 * e^(2/3)) / (4 * e - 4 * e^(1/2)) at x=3/2.[/FONT]

 

[Ishuda]

Hello,
since two days I'm trying to get trough it but somehow always stuck at the same point during the process.

differential equation is: x^3*y" + x*y' + d*y = 0

which has an irregular singular point at x=0 (I know how to prove it), and even though it is an irregular point a solution is possible to obtain using Frobenius method of the Frobenius form: y=sum_(n=0)^∞ (a_n * x^(n+v))

...
I have to show that the particular solution is y= (3 * e - 3 * e^(2/3)) / (4 * e - 4 * e^(1/2)) at x=3/2.

The problem is that the Frobenius method is for regular singular points and you have an equation with an irregular singular point. For irregular singular points, the method is somewhat different, see for example
http://ocw.mit.edu/courses/mathemat...ngineering-fall-2004/lecture-notes/eight1.pdf
BTW, the problem the paper is working on for a second order ODE with an irregular singular point is
y'' + c(x) y' + d(x) y = 0
which is obtained from the previous lecture
http://ocw.mit.edu/courses/mathemat...ngineering-fall-2004/lecture-notes/seven1.pdf

 

[matematicar73]

The problem is that the Frobenius method is for regular singular points and you have an equation with an irregular singular point. For irregular singular points, the method is somewhat different, see for example
http://ocw.mit.edu/courses/mathemat...ngineering-fall-2004/lecture-notes/eight1.pdf
BTW, the problem the paper is working on for a second order ODE with an irregular singular point is
y'' + c(x) y' + d(x) y = 0
which is obtained from the previous lecture
http://ocw.mit.edu/courses/mathemat...ngineering-fall-2004/lecture-notes/seven1.pdf

This both is a great reading. Thank you very much.
But still I am stuck with my exercise. Maybe I should exactly state the exercise. It says:
Show that the ordinary differential equation x^3y" + xy' + dy = 0, where d is a constant, has an irregular singular point at x=0. By considering a solution of the Frobenius form y=sum_(n=0)^[FONT=Helvetica Neue, Helvetica, Arial, sans-serif]∞ (a_n)x^(n+v), show that only one value of v is possible. Given d = -1, 0, 1, 2, ...., show that the series solution terminates, producing a solution consisting of a finite number of terms.

To show that it is an irregular point at x=0 is easy. I've done that! [/FONT][FONT=Helvetica Neue, Helvetica, Arial, sans-serif]
But here is where I'm stuck. I am not getting anything since I am going from the fact that a_0 does not equal zero but I always get that exactly a_0=0.
How can I determine v?

After that I got: d = -1 and I should find general solution and then the particular solution which should outcome with the fact given already: y=(3e - 3e^(2/3)) / (4e - 4e^(1/2)) for x=3/2

But I can't simply get the start of determine v. [/FONT]