y' - 4y = e^2x who can help me with this?

Question

its a linear differential equation

Answer 1

Given y'' - 4y = e^2x

Rewrite the equation
y'' - 4y = ℮²x / x
y'' - 4y = e²

Find the complementary function by solving the auxiliary equation:
m² - 4 = 0
m² = 4
m = ±2
yᶜ = A℮^(2x) + B℮^(-2x)
yᶜ = A℮^(2x) + B / ℮^(2x)

Find the particular integral by comparing coefficients:
yᵖ = C
yᵖ' = 0
yᵖ'' = 0
yᵖ'' - 4yᵖ = ℮²
-4C = ℮²
4C = -℮²
C = -℮² / 4

Find the general solution by combining these two parts:
y = yᶜ + yᵖ
y = A℮^(2x) + B / ℮^(2x) - ℮² / 4

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