Question

Which of the following functions are solutions of the differential equation y+4y+4y=0? A. y(x) =e^-2x B. y(x) = e^+22 C. y(x) = xe^-2x D. y(x) = -2x E. y(x) = 0 F. g(x) = x^2e^-2x

Answer 1

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.