matematicar73
New member
- Joined
- Dec 16, 2015
- Messages
- 8
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]
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]