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