The differential eqn for y'' = -y HELP PLEASE :(

I understand that it is an imaginary number, and the roots are the same. How to i solve?

The roots are not the same, the roots are +i and -i, proceed as usual.

y = A' e^{it} + B' e^{-it}, A' and B' = constants

= (A' + B') cos(t) + i(A' - B') sin(t) <---- From the equality e^{it} = cos(t) + i sin(t) as per Euler

= A cos(t) + B sin(t)

where we define A = A' + B', and B = i(A' - B')

Since A' and B' are just constants, A = A' + B' represents just sum of two numbers, you can add them together without consequence and relabel the constant as a single term A = A' + B'

Say the constant A = 5, now we defined A as A = A' + B', so A' + B' = 5, maybe A' = 2, B' = 3, maybe A' =-1, B' = 6

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The point is, it does not matter what A' and B' are, they both add together to give a single cos(t) term, so we can write A = 5 and still have all the information about the cos(t) term without having to fuss with what A' and B' are individually. I just relabeled the constants as A and B

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i understand now. thank you so much!

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y'' = -y

y'' + y = 0

Substitute y = e^(rx).

r^2 + 1 = 0

r^2 = i^2

r = +/- i

y = A*sin(x) + B*cos(x)