(x+2)y''+(2x+5)y'+2y=(x+1)eˣ.
Let y=a₀+a₁x+a₂x²+...=∑aᵣxʳfor r∈[0,∞)=a₀+a₁x+∑aᵣ₊₂xʳ⁺².
So y'=∑raᵣxʳ⁻¹=a₁+∑(r+2)aᵣ₊₂xʳ⁺¹ and y''=∑(r+2)(r+1)aᵣ₊₂xʳ.
eˣ=1+x+∑xʳ⁺²/(r+2)!.
(x+1)eˣ=(x+1)(1+x+∑xʳ⁺²/(r+2)!)=x²+2x+1+(x+1)(∑xʳ⁺²/(r+2)!).
All the summations applies for r∈[0,∞).
The DE can be written:
(x+2)∑(r+2)(r+1)aᵣ₊₂xʳ+
(2x+5)(a₁+∑(r+2)aᵣ₊₂xʳ⁺¹)+
2(a₀+a₁x+∑aᵣ₊₂xʳ⁺²)=
x²+2x+1+(x+1)(∑xʳ⁺²/(r+2)!).
This can be further expanded:
∑(r+2)(r+1)aᵣ₊₂xʳ⁺¹+2∑(r+2)(r+1)aᵣ₊₂xʳ+
2a₁x+5a₁+2∑(r+2)aᵣ₊₂xʳ⁺²+5∑(r+2)aᵣ₊₂xʳ⁺¹+
2a₀+2a₁x+2∑aᵣ₊₂xʳ⁺²=
x²+2x+1+∑xʳ⁺³/(r+2)!+∑xʳ⁺²/(r+2)!.
Next, group together the same powers of x:
∑2(r+3)aᵣ₊₂xʳ⁺²+
∑(r²+8r+12)aᵣ₊₂xʳ⁺¹+
2∑(r+2)(r+1)aᵣ₊₂xʳ+
4a₁x+
5a₁+2a₀=
x²+2x+1+∑xʳ⁺³/(r+2)!+∑xʳ⁺²/(r+2)!.
Now we need to match terms with the same power of x. To do this accurately we need to find which r values to substitute in each of these summation groups. If we start at x⁰ (constant terms), the only contributor is constant (x⁰):
2∑(r+2)(r+1)aᵣ₊₂xʳ+5a₁+2a₀=1,
Putting r=0, we get:
4a₂+5a₁+2a₀.
The contributing groups for x¹ are:
∑(r²+8r+12)aᵣ₊₂xʳ⁺¹ (r=0) and 2∑(r+2)(r+1)aᵣ₊₂xʳ(r=1) and 4a₁x.
This gives us:
12a₂+12a₃+4a₁.
For x²:
2∑(r+3)aᵣ₊₂xʳ⁺² (r=0) +
∑(r²+8r+12)aᵣ₊₂xʳ⁺¹ (r=1) +
2∑(r+2)(r+1)aᵣ₊₂xʳ(r=2).
This gives us:
6a₂+21a₃+24a₄.
For x³:
2∑(r+3)aᵣ₊₂xʳ⁺² (r=1) +
∑(r²+8r+12)aᵣ₊₂xʳ⁺¹ (r=2) +
2∑(r+2)(r+1)aᵣ₊₂xʳ(r=3).
This gives us:
8a₃+32a₄+40a₅.
On the right-hand side we have:
1+2x+x²+x²/2!+x³/2!+x³/3!+...
Equating coefficients:
x⁰: 2a₀+5a₁+4a₂=1.
x¹: 4a₁+12a₂+12a₃=2, 2a₁+6a₂+6a₃=1.
x²: 6a₂+21a₃+24a₄=1+½=3/2, 4a₂+14a₃+16a₄=1.
x³: 8a₃+32a₄+40a₅=½+⅙=⅔, 12a₃+48a₄+60a₅=1.
For xʳ:
2(r+1)aᵣ+(r+1)(r+5)aᵣ₊₁+2(r+1)(r+2)aᵣ₊₂=(r+1)/(r-1)! where r≥0. Note that factorial for values ≤ 0 is understood to be 1. Hence, for example, (-1)!=0!=1.
This formula relates the coefficients for each power r of x, so can be used to determine any coefficient from its predecessors.
We know there are going to be 2 constants of integration. If we make a₀ and a₁ those two constants A and B respectively, then we should be able to relate other coefficient to A and B.
a₀=A, a₁=B,
a₂=¼(1-2A-5B).
a₃=(6A+11B-1)/12.
Therefore the solution up to x³ is:
y=A+Bx+(1-2A-5B)x²/4+(6A+11B-1)x³/12+...