Differential Equations Problem please help?

I do not know that method, but I am sure it is not the simplest way.

Find the complementary function by solving the auxiliary equation:

x'' + 2x' + x = 0

m² + 2m + 1 = 0

(m + 1)² = 0

m + 1 = 0

m = -1

xᶜ = (At + B)℮⁻ᵗ

Find the particular integral by comparing coefficients:

xᵖ = C℮^(t + 1)

xᵖ' = C℮^(t + 1)

xᵖ'' = C℮^(t + 1)

xᵖ'' + 2xᵖ' + xᵖ = ℮^(t + 1)

C℮^(t + 1) + 2C℮^(t + 1) + C℮^(t + 1) = ℮^(t + 1)

4C℮^(t + 1) = ℮^(t + 1)

4C = 1

C = ¼

xᵖ = ℮^(t + 1) / 4

Find the general solution by combining these two parts:

x = xᶜ + xᵖ

x = (At + B)℮⁻ᵗ + ℮^(t + 1) / 4

x = ℮^(t + 1) / 4 + (At + B)℮⁻ᵗ

Find the particular solution by solving for the constants:

When t = 0, x = 0

℮ / 4 + B = 0

B = -℮ / 4

x' = ℮^(t + 1) / 4 - [At - (A - B)]℮⁻ᵗ

When t = 0, x' = 0

℮ / 4 + A - B = 0

A = B - ℮ / 4

A = -℮ / 4 - ℮ / 4

A = -℮ / 2

x = ℮^(t + 1) / 4 + (-℮t / 2 - ℮ / 4)℮⁻ᵗ

x = ℮^(t + 1) / 4 - (2℮t + ℮)℮⁻ᵗ / 4

x = ℮^(t + 1) / 4 - (2t + 1)℮℮⁻ᵗ / 4

x = ℮^(t + 1) / 4 - (2t + 1)℮^(1 - t) / 4

x = ℮^(t + 1) / 4 - (2t + 1) / [4℮^(t - 1)]