6. Consider the differential equation 2xy" - 4y' - y = 0. i) Show that x=0 is a regular singular point, ii) Find a series solution about x=0 corresponding to the larger root of the indicial equation.

Question

please help me 

Answer 1

The DE is 2xy"-4y'-y=0.
Let y=∑a_nxⁿ, then y'=∑na_nx^n-1 and y"=∑n(n-1)a_nx^n-2. The limits of the summation are 0≤n<∞.
These represent y=a₀+a₁x+a₂x²+a₃x³+a₄x⁴+a₅x⁵+...
y'=a₁+2a₂x+3a₃x²+4a₄x³+5a₅x⁴+...
y"=2a₂+6a₃x+12a₄x²+20a₅x³+...
The DE becomes:
(1) 2∑n(n-1)a_nx^n-1-4∑na_nx^n-1-∑a_nxⁿ=0,
(2) ∑(2n(n-1)a_nx^n-1-4na_nx^n-1)-∑a_nxⁿ=0.
(3) ∑2n(n-3)a_nx^n-1-∑a_nxⁿ=0,
(4) ∑2n(n-3)a_nx^n-1-∑a_nxⁿ-a₀=0, where summation limits are now 1≤n<∞.
However, we need to consider the first few terms for n<3 before this will make sense, because n-3 would be negative. Let's see what happens when we expand the summation for the above equations.
If we use the expanded terms, the DE becomes:
(5) 2x(2a₂+6a₃x+12a₄x²+20a₅x³+...)
-4(a₁+2a₂x+3a₃x²+4a₄x³+5a₅x⁴+...)
-(a₀+a₁x+a₂x²+a₃x³+a₄x⁴+a₅x⁵+...)=0.
From these we can create some more equations, if we put x=0:
(6) -4a₁-a₀=0, so a₁=-¼a₀, making (5):
(7)  2x(2a₂+6a₃x+12a₄x²+20a₅x³+...)

-4(-¼a₀+2a₂x+3a₃x²+4a₄x³+5a₅x⁴+...)

-(a₀-a₀x/4+a₂x²+a₃x³+a₄x⁴+a₅x⁵+...)=0.

This becomes:

(8) x(-4a₂+a₀/4)+x²(-a₂)+x³(8a₄-a₃)+...=0. Note that the succession of terms denoted by the ellipsis contain powers of x greater than 3.

We can divide through by common factor x, because in the general case x≠0:

(9) -4a₂+a₀/4-a₂x+x²(8a₄-a₃)+...=0. When x=0, we get:

-4a₂+a₀/4=0, making a₂=a₀/16. This enables us to write (9):

(10) -a₀x/16+x²(8a₄-a₃)+...=0.

Again, x is a common factor, so:

-a₀/16+x(8a₄-a₃)+...=0. This implies that a₀=0, which makes a₁=a₂=0.

We can write y=a₃x³+a₄x⁴+a₅x⁵+... because there is no constant term nor x or x2 terms. (i) Therefore y(0)=0, y'(0)=0 and y"(0)=0, a regular single point for y and its two derivatives (expressed in the DE). This leads to a different way of expressing y, as we shall see later.

Now we turn back to equation 4, which now becomes:

(11) ∑2n(n-3)a_nx^n-1-∑a_nxⁿ=0. We know that the polynomial for y starts with a₃x³, so we can plug n=3, 4, etc., into this equation:

We get -a₃x³+8a₄x³-a₄x⁴+20a₅x⁴-a₅x⁵+36a₆x⁵-a₆x⁶+...=0,

x³(8a₄-a₃)+x⁴(20a₅-a₄)+x⁵(36a₆-a₅)+...=0.

a₄=a₃/8, a₅=a₄/20=a₃/160, a₆=a₅/36=a₃/5760, ..., because each coefficient must be zero for general x.

Each coefficient is expressible in terms of one arbitrary constant a₃. We can re-label this a₀. Now y can be expressed:

y=a₀x³(1+x/8+x²/160+x³/5760+...).

I guess that from this the formula given can be derived. Certainly the coefficients I calculated are correct according to the formula. I'm a little puzzled as to why a₀ is only applied to the summation so that there is a lone x³ term without the a₀ coefficient. But perhaps there's a mistake in my solution. I hope my solution helps.