Let F=xy"+y'+xy=0.
A power series for x can be written:
y=a0+a1x+a2x^2+a3x^3+a4x^4+...+a[r]x^r+...; or y=∑a[r]x^r for 0≤r≤∞ or y=∑a[r]x^r{0,∞}.
y'=a1+2a2x+3a3x^2+4a4x^3+...+ra[r]x^(r-1)+...; or y'=∑ra[r]x^(r-1){0,∞};
y"=2a2+6a3x+12a4x^2+...+r(r-1)a[r]x^(r-2)+...; or y"=∑r(r-1)a[r]x^(r-2){0,∞};
Substituting for y and its derivatives in F:
xy"=2a2x+6a3x^2+12a4x^3+...+r(r-1)a[r]x^(r-1)+...; or xy"=∑r(r-1)a[r]x^(r-1){0,∞};
y'=a1+2a2x+3a3x^2+4a4x^3+...+ra[r]x^(r-1)+...; or y'=∑ra[r]x^(r-1){0,∞};
xy=a0x+a1x^2+a2x^3+a3x^4+a4x^5+...+a[r]x^(r+1)+...; or y=∑a[r]x^r(r+1) for 0≤r≤∞ or y=∑a[r]x^r{0,∞}.
We can add xy" to y' combining like powers of x: xy"+y'=∑(r(r-1)a[r]x^(r-1)+ra[r]x^(r-1)){0,∞}.
This simplifies because ra[r]x^(r-1) term cancels out and we get xy"+y'=∑r^2a[r]x^(r-1){0,∞}.
To add in the last term xy so that the same power of x is used we have to step back two terms. Let's see what happens if we expand the summation:
xy"+y'+xy=a1+x(a0+4a2)+x^2(a1+9a3)+x^3(a2+16a4)+x^4(a3+25a5)+...
A pattern emerges so we can write F=a1+∑x^(r-1)(a[r-2]+r^2a[r]{2,∞}=0.
The terms have to sum to zero, so a1=0, implying that a3, a5, etc.=0.
Now we need to find a relationship between a0, a2, a4, etc.
a[r-2]+r^2a[r]=0, so a[r]=-a[r-2]/r^2. When r=2, we get a2=-a0/2^2.
From this it follows that, putting r=4, we get a4=-a2/4^2=(-1/4^2)(-1/2^2)a0=a0/((2^2)(4^2)).
And when r=6: a6=-a0/((2^2)(4^2)(6^2)). a8=a0/((2^2)(4^2)(6^2)(8^2)). We can see a pattern emerging.
(2^2)(4^2)(6^2)(8^2)=(2*4*6*8)^2=(1*2*3*4)^2 * (2^2)^4. If r=2n, then in general we get ((n!)^2)(2^2n).
[CHECK: n=1, (1^2)(2^2)=4; n=2, (2^2)(2^4)=64; n=3, (6^2)(2^6)=2304.]
But terms have alternating signs. When n=1, the sign is negative, when n=2 it's positive so we introduce the factor (-1)^n to get the right sign. Write a0 as constant a, and then all other terms are multiples of a (effectively a constant of integration).
For example, a2=-a/4, a4=-a2/16=a/64, a6=-a4/36=-a/2304, etc.
Therefore y=a(1-x^2/4+x^4/64-x^6/2304+...+(-1)^n.x^(2n)/((n!)^2)(2^2n))+...).