For what values of r does the function
y = erx
satisfy the differential equation
y'' − 8y' + 2y = 0?
(Enter your answers as a comma-separated list.)
To find the values of r that satisfy the given differential equation, we need to find the characteristic equation of the differential equation by substituting y = e^(rx) into the equation.
The characteristic equation is obtained by substituting y = e^(rx) into the differential equation:
y'' - 8y' + 2y = 0
Differentiating y twice, we get:
y' = re^(rx)
y'' = r^2e^(rx)
Substituting these results back into the differential equation:
r^2e^(rx) - 8re^(rx) + 2e^(rx) = 0
Factoring out e^(rx):
e^(rx)(r^2 - 8r + 2) = 0
Since e^(rx) is never zero, we can ignore it and focus on solving the quadratic equation:
r^2 - 8r + 2 = 0
To solve this quadratic equation, we can use the quadratic formula:
r = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -8, and c = 2. Substituting these values into the quadratic formula:
r = (-(-8) ± sqrt((-8)^2 - 4(1)(2))) / (2(1))
= (8 ± sqrt(64 - 8)) / 2
= (8 ± sqrt(56)) / 2
= (8 ± 2sqrt(14)) / 2
= 4 ± sqrt(14)
Thus, the values of r that satisfy the differential equation are:
r = 4 + sqrt(14), 4 - sqrt(14)
Hence, the solution is r = 4 + sqrt(14), 4 - sqrt(14).