Show that both functions y1=e^(-2x) and y2=xe^(-2x) are solutions to the differential equation y^''+4y^'+4y=0

Answer 1

what's the troub;e? Just plug and chug

y1 = e^(-2x)
y1' = -2e^(-2x)
y1" = 4e^(-2x)

y"+4y'+4 = 0

y2 = xe^(-2x)
y2' = (1-2x) e^(-2x)
y2" = (4x-4) e^(-2x)

y"+4y'+4y = 0