Question

Differential eqn: solve y"=-w^2y

Answer 1

are you considering w as constant or variable?

Answer 2

hmmm...probably a variable

Answer 3

sorry, constant

Answer 4

its been a long day lol

Answer 5

1/((2lnw)^2)*w^(-2y)

Answer 6

wow, I don't remember that at all. Did you use LaPlace transform?

Answer 7

y" +w^2 y= 0 w constant
the characteristic equation is
m^2 + w^2= 0
so m= ±w
and
\[ y= c_1 e^{wt}+c_2 e^{-wt} \]