Let’s just solve first.
We can rewrite:
y"-2y'+y+y'/x-y/x=0.
If we can find y=f(x) such that (1) y"-2y'+y=0 and (2) y'/x-y/x=0 simultaneously, then we have found one solution to the differential equation.
The characteristic equation for (1) gives us f(x)=Axeˣ+Beˣ.
The solution for (2) is y'=y, that is, f(x)=aeˣ.
If we reconcile these by a=B, A=0, then y=f(x)=Beˣ.
Going back to the original DE, substitute for y, noting that y'=y"=y. Therefore:
xy+(1-2x)y-(1-x)y=xy+y-2xy-y+xy=0, so y=Beˣis a general solution. But see what we get using Frobenius’ Method.
FROBENIUS
y=∑(a_n)x^(n+r) for 0≤n≤∞ where a_n is “a sub n”, a constant coefficient for each x term.
r is given by the indicial equation r(r-1)+p₀r+q₀=0. To find p₀ and q₀ we first divide the DE by x:
y"+(1-2x)y'/x-(1-x)y/x=0 (x≠0).
Define P(x)=(1-2x)/x, Q(x)=(1-x)/x, and
p(x)=xP(x)=1-2x, q(x)=x²Q(x)=x(1-x), p₀ is defined as p(0)=1 and q₀ as q(0)=0.
So, r(r-1)+r=0=r² and r=0. Therefore y=∑(a_n)x^(n+r)=∑(a_n)xⁿ.
Therefore y'=∑(a_n)nxⁿ⁻¹ and y"=∑(a_n)n(n-1)xⁿ⁻².
Now substitute this series in the original DE:
x∑(a_n)n(n-1)xⁿ⁻²+(1-2x)∑(a_n)nxⁿ⁻¹-(1-x)∑(a_n)xⁿ=
∑(a_n)n(n-1)xⁿ⁻¹+∑(a_n)nxⁿ⁻¹-2∑(a_n)nxⁿ-∑(a_n)xⁿ+∑(a_n)xⁿ⁺¹=0,
∑(a_n)xⁿ⁻¹[n²-n+n]-∑(a_n)xⁿ(2n+1)+∑(a_n)xⁿ⁺¹=0,
∑(a_n)n²xⁿ⁻¹-∑(a_n)xⁿ(2n+1)+∑(a_n)xⁿ⁺¹=0 for 0≤n≤∞.
Expand the first few terms:
a₁+4a₂x+9a₃x²+16a₄x³+...
-(a₀+3a₁x+5a₂x²+7a₃x³+...)
+a₀x+a₁x²+a₂x³+a₃x⁴+a₄x⁵+...=0.
Now group according to the power of x:
(a₁-a₀)+x(4a₂-3a₁+a₀)+x²(9a₃-5a₂+a₁)+x³(16a₄-7a₃+a₂)+...=0.
Each coefficient must be zero to satisfy general x.
Therefore a₁=a₀; 4a₂-3a₁+a₀=0, 4a₂=2a₀, a₂=a₀/2; 9a₃-5a₀/2+a₀=0, 9a₃=3a₀/2, a₃=a₀/6; 16a₄=7a₃-a₂=7a₀/6-a₀/2=2a₀/3, a₄=2a₀/48=a₀/24; 25a₅=9a₄-a₃=3a₀/8-a₀/6=5a₀/24, a₅=a₀/120; ...
This gives us y=a₀(1+x+x²/2+x³/6+x⁴/24+x⁵/120+...+xⁿ/n!+...)=a₀eˣ.
This is the same result as we got using a different method.