The DE is 2xy"-4y'-y=0.
Let y=∑anxn, then y'=∑nanxn-1 and y"=∑n(n-1)anxn-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)anxn-1-4∑nanxn-1-∑anxn=0,
(2) ∑(2n(n-1)anxn-1-4nanxn-1)-∑anxn=0.
(3) ∑2n(n-3)anxn-1-∑anxn=0,
(4) ∑2n(n-3)anxn-1-∑anxn-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)anxn-1-∑anxn=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.