solve initial value problem y"+9y=0,y(pi)=2,y'(pi)=3

Question

initial value problem

Answer 1

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)