Find the general solution of the given secondorder different

Find the general solution of the given second-order differential equation
y\'\' + 4y\' + 4y = 0
the answer is c_1(e^-2x) + c_2(e^-2x)x but i don\'t know where the other x came from in c_2

Solution

y\'\' + 4y\' + 4 = 0

Let y\' = m...

m^2 + 4m + 4 = 0

(m +2)(m + 2) = 0

m = -2 , -2

Since these are repeated roots, the result is of the form :

c1*e^(root * x) + c2 * x * e^(root * x)

Here root is -2

So, plug that in :

y = c1*e^(-2x) + c2*x*e^(-2x) ---> ANSWER