Differentiate the function. Which of the following functions satisfy y^n + y = 0?

Question

Which of the following functions satisfy y^n + y = 0?
a. y1 = 2sin x + 3 cos x
b. y2 = 4 sin x - (pie)cos x
c. y3 = x sin x

Simplify answers and please show work
I'M REALLY TRYING TO GRASP THIS CONCEPT

y^n IS y raised to the n i guess same as 2^2

Answer 1

I think in this case y^n is referring to the n'th derivative of y, rather than the n'th power of y. Often this is written as y(n) (with the (n) superscripted like a power) to avoid confusion with y^n.

So, for (a) we have y = 2 sin x + 3 cos x, y' = 2 cos x - 3 sin x, y'' = -2 sin x - 3 cos x = -y, so y'' + y = 0.

Similarly, for (b) we have y = 4 sin x - π cos x, y' = 4 cos x + π sin x, y'' = -4 sin x + π cos x = - y, so again y'' + y = 0.

For (c) we have y = x sin x, y' = sin x + x cos x, y'' = 2 cos x - x sin x. From this it follows that y(4) = - 4 cos x + x sin x, y(6) = 6 cos x - x sin x, etc. No matter how often we differentiate we won't get back to -x sin x. Odd number derivatives will have a sin x term and an x cos x term, while even numbered derivatives will have a cos x term and an x sin x term.

So y1 and y2 satisfy y'' + y = 0, but y3 does not satisfy y(n) + y = 0 for any positive integer n.