Differential Equation: Need Verification Please

Question

Determine the specific solution of the following initial-value problem. y''+ 4y = 0, y(pi)=0 and y'(pi)=1

My answer is: y=(0)cos(2x) + C₂sin(2x) or simply y=C₂sin(2x) Note: the "(0)" is my C1. I couldn't find the value for my C₂.

Answer 1

y'' + 4y = 0

 

General solution:

 

y = a cos(2x) + b sin(2x)

 

y(π) = 0 ⇒

 

0 = a cos(2π) + b sin(2π) = a + 0

a = 0

 

y = b sin(2x)

 

y'(π) = 1 ⇒

 

2b cos(2π) = 1

 

2b = 1

b = 1/2

 

y(x) = (1/2)sin(2x)

 

Your C₂ = 1/2

 

Test the solution against the given boundary conditions.