Question

Show that the function y=Ae^-x + Be^-x satisfies the differential equation y"+2y'+y=0

Answer 1

y = A(e^-x) + B(e^-x)
y' = -A(e^-x) - B(e^-x) = -1[A(e^-x) + B(e^-x)] = -y
y" = A(e^-x) + B(e^-x) = +y

y = +y
y' = -y
y" = +y

y" + 2y' + y = 0
(+y) + 2(-y) + (+y) = 0
y - 2y + y = 0
2y - 2y = 0
0y = 0
0 = 0

Or, the long way...

y" + 2y' + y = 0
(A(e^-x) + B(e^-x)) + 2(-A(e^-x) - B(e^-x)) + (A(e^-x) + B(e^-x)) = 0
2A(e^-x) + 2B(e^-x) - 2A(e^-x) - 2B(e^-x) = 0
0A(e^-x) + 0B(e^-x) = 0
0 = 0

Therefore, y = A(e^-x) + B(e^-x) does satisfy the differential equation
y" + 2y' + y = 0.