Question: solve initial value problem y"+9y=0,y(pi)=2,y'(pi)=3
Diff eqn: y"+9y=0.
Aux eqn: m^2 + 9 = 0
i.e. m = +/- 3i
Hence, genl. soln is y(x) = A.cos(3x) + B.sin(3x)
Also, y'(x) = -3A.sin(3x) + 3B.cos(3x)
Initial conditions
y(pi) = 2 = A.cos(3pi) + B.sin(3pi) = -A + 0 => A = -2
y'(pi) = 3 = -3A.sin(3pi) + 3B.cos(3pi) = 0 - 3B => B = -1
Solution: -2.cos(3x) - sin(3x)