To find the general solution to the given differential equation:
y" - 4y' + 8y = 0
We can start by assuming a solution of the form y = e^(rt), where r is a constant. Substituting this into the differential equation, we get:
r^2e^(rt) - 4re^(rt) + 8e^(rt) = 0
Factoring out e^(rt), we have:
e^(rt)(r^2 - 4r + 8) = 0
For this equation to hold, either e^(rt) = 0 (which is not possible) or r^2 - 4r + 8 = 0. Solving the quadratic equation, we find the roots:
r = (4 ± √(4^2 - 4 * 1 * 8)) / (2 * 1)
r = (4 ± √(-16)) / 2
r = (4 ± 4i) / 2
r = 2 ± 2i
The roots are complex, so we have:
r1 = 2 + 2i
r2 = 2 - 2i
Since the roots are complex conjugates, we can write the general solution as:
y = c1e^(2t)cos(2t) + c2e^(2t)sin(2t)
where c1 and c2 are arbitrary constants.
Therefore, the general solution to the given differential equation is:
y = c1e^(2t)cos(2t) + c2e^(2t)sin(2t)