Answer:
The general solution: [tex]C_{1}e^{-4x} + xC_{2}e^{-4x}[/tex]
Step-by-step explanation:
Differential equation: y'' + 8y' + 16y = 0
We have to find the general solution of the above differential equation.
The auxiliary equation for the above equation can be writtwn as:
m² + 8m +16 = 0
We solve the above equation for m.
(m+4)² = 0
[tex]m_{1}[/tex] = -4, [tex]m_{2}[/tex] = -4
Thus we have repeated roots for the auxiliary equation.
Thus, the general solution will be given by:
y = [tex]C_{1}e^{m_{1}x} + xC_{2}e^{m_{2}x}[/tex]
y = [tex]C_{1}e^{-4x} + xC_{2}e^{-4x}[/tex]
