Solution about regular singular point For the following diff

Solution about regular singular point For the following differential equation 2xy\" + y\' + xy = 0 Show that the given differential equation has a regular singular point at x = 0. Determine the indicial equation, the recurrence relation, and the squareroots of the indicial equation. Find the series solution (x > 0) corresponding to the larger squareroot. If the squareroots are unequal and do not differ by an integer, find the series solution corresponding to the smaller squareroot also.

Solution

Answer (a)

Given equation is (2x)(d2y/dx2) + (1)(dy/dx) + (x)y = 0

                          (d2y/dx2) + (1/2x)(dy/dx) + (x/2x)y = 0

So, p(x) = 1/(2x) and q(x) = 1 / 2

Here p(x) approaches ± as x approaches 0. So the given differential equation has a regular singular point at x = 0

Answer (b)

Assume that y = (n = 0 to ) a(n) x^(n+r)

Substitute this into the Differential Equation

    = 2x(n=0 to )(n+r)(n+r-1)a(n)x^(n+r-2)

           + (n=0 to )(n+r)a(n)x^(n+r-1)

                + x (n = 0 to ) a(n) x^(n+r) = 0

Simplifying:
    = (n = 0 to ) (n+r) [2(n+r-1) + 1] a(n) x^(n+r-1)

               + (n = 0 to ) a(n) x^(n+r+1) = 0.

Re-index to x^(n+r-1):
     = (n = 0 to ) (n+r)(2n+2r-1) a(n) x^(n+r-1)

                 + (n = 2 to ) a(n-2) x^(n+r-1) = 0.

Collecting like terms:
     = r(2r-1) a(0) x^(r-1) + (r+1)(2r+1) a(1) x^r

             + (n = 2 to ) [(n+r)(2n+2r-1) a(n) + a(n-2)] x^(n+r-1) = 0.

Setting the first coefficient equal to 0 yields the indicial equation r(2r-1) = 0 ==> r = 1/2.

Answer (c)

Now, we find the solution corresponding to r = 1/2:

           3 a(1) x^r + (n = 2 to ) [n(2n+1) a(n) + a(n-2)] x^(n+1/2) = 0.

Equating like coefficients (with a(0) arbitrary):
           3 a(1) = 0 ==> a(1) = 0

For n = 2, 3, ...:
n(2n+1) a(n) + a(n-2) = 0
==> a(n) = -a(n-2)/(n(2n+1)).

Since a(1) = 0, we have that a(3) = a(5) = a(7) = ... = 0.

For even n:
       a(2) = -a(0)/(2 * 5) = -a(0)/10
       a(4) = -a(2)/(4 * 9) = a(2)/360
       ...

So, a solution for r = 1/2 is
y = a(0) x^(1/2) [1 + (1/10)x^2 + (1/360)x^4 + ...]

Answer (d)

Here roots are equal and differ by an integer.