y = a e^kx
y' = a k e^kx
y" = a k^2 e^kx
4 a k^2 = 9 a
k = +/- 3/2
more interesting would be
4 y" + 9 y = 0
then you would get
k^2 = -9
and k = 1.5 i
so
y = a e^1.5 i x
but e^ix = cos x + i sin x
Find the complete general solution to the second order DE:
4y'' - 9y = 0
3 answers
y = a e^kx
y' = a k e^kx
y" = a k^2 e^kx
4 a k^2 = 9 a
k = +/- 3/2
more interesting would be
4 y" + 9 y = 0
then you would get
k^2 = -9/4
and k = 1.5 i
so
y = a e^1.5 i x
but e^ix = cos x + i sin x
y' = a k e^kx
y" = a k^2 e^kx
4 a k^2 = 9 a
k = +/- 3/2
more interesting would be
4 y" + 9 y = 0
then you would get
k^2 = -9/4
and k = 1.5 i
so
y = a e^1.5 i x
but e^ix = cos x + i sin x
you know it will be some exponentials e^ax where the exponents solve 4a^2-9 = 0
y = c1 e^(3/2 x) + c2 e^(-3/2 x)
y = c1 e^(3/2 x) + c2 e^(-3/2 x)