(D²-3D+2)y=e^eˣ.
The characteristic equation gives us a partial solution.
D²-3D+2 “factorises” into (D-1)((D-2), so this partial solution is y=Aeˣ+Be²ˣ.
Now we need to find a particular solution for e^eˣ.
We can write this as a series:
1+eˣ+e²ˣ/2!+e³ˣ/3!+...+eʳˣ/r!=∑eʳˣ/r! for integer r≥0.
We can use the additive property of differentiation to find a solution for each term in this series. For example, when r=0, we can find y₀ such that:
D²y₀-3Dy₀+2y₀=1, where D=d/dx and D²=d²/dx².
For r=0, the solution is y₀=a₀, where a₀ is a constant, so D=D²=0 and 2y₀=2a₀=1, making a₀=½.
We can write the solution as:
y=Aeˣ+Be²ˣ+y₀+y₁+y₂+...+yᵣ where yᵣ is the solution for eʳˣ/r!.
For r=1, y₁=a₁xeˣ. We cannot use a₁eˣ because we already have Aeˣ.
Dy₁=a₁xeˣ+a₁eˣ, D²y₁=a₁xeˣ+2a₁eˣ.
D²y₁-3Dy₁+2y₁=
a₁xeˣ+2a₁eˣ-3(a₁xeˣ+a₁eˣ)+2a₁xeˣ=
-a₁eˣ≡eˣ, making a₁=-1.
For r=2, y₂=a₂xe²ˣ, since we can’t use a₂e²ˣ because of Be²ˣ.
Dy₂=2a₂xe²ˣ+a₂e²ˣ, D²y₂=4a₂xe²ˣ+2a₂e²ˣ+2a₂e²ˣ.
D²y₂-3Dy₂+2y₂=
4a₂xe²ˣ+2a₂e²ˣ+2a₂e²ˣ-3(2a₂xe²ˣ+a₂e²ˣ)+2a₂xe²ˣ=
a₂e²ˣ≡e²ˣ/2!, making a₂=½.
So far we have y=Aeˣ+Be²ˣ+½-xeˣ+½xe²ˣ.
For r>2, yᵣ=aᵣeʳˣ, Dyᵣ=raᵣeʳˣ, D²yᵣ=r²aᵣeʳˣ.
D²yᵣ-3Dyᵣ+2yᵣ=r²aᵣeʳˣ-3raᵣeʳˣ+2aᵣeʳˣ≡eʳˣ/r!.
Therefore aᵣ=1/(r!(r²-3r+2))
and y=Aeˣ+Be²ˣ+½-xeˣ+½xe²ˣ+...+eʳˣ/(r!(r²-3r+2)), for integer r>2, is the particular solution.