Calculus
posted by Sara on .
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' = (12x) e^(2x)
y2" = (4x4) e^(2x)
y"+4y'+4y = 0