Answer: E. y(x) = 0
Step-by-step explanation:
y(x) = 0 is the only answer from the options that satisfies the differential equal y" - 4y' + 4y = 0
See:
Suppose y = e^(-2x)
Differentiate y once to have
y' = -2e^(-2x)
Differentiate the 2nd time to have
y" = 4e^(-2x)
Now substitute the values of y, y', and y" into the give differential equation, we have
4e^(-2x) - 4[-2e^(-2x)] + 4e^(-2x)
= 4e^(-2x) + 8e^(-2x) + 4e^(-2x)
= 16e^(-2x)
≠ 0
Whereas we need a solution that makes the differential equation to be equal to 0.
If you test for the remaining results, the only one that gives 0 is 0 itself, and that makes it the only possible solution from the options.
It is worth mentioning that apart from the trivial solution, 0, there is a nontrivial solution, but isn't required here.
