Solve the given initial-value problem in which the input function g(x) is discontinuous. [Hint: Solve the problem on two intervals, and then find a solution so that y and y' are continuous at x = π/2.] y'' + 4y = g(x), y(0) = 1, y'(0) = 4, where g(x) = sin(x), 0 ≤ x ≤ π/2 0, x > π/2

(1)

Therefore the complete solution is

(2)

Therefore the complete solution is

Step-by-step explanation:

Given differential equation is

y''+4y = g(x) is associated  with y''+4y=0.

The auxiliary equation is

m²+4=0

⇒m²= -4

The complimentary function is

Case 1:

g(x)= sinx

Then y''+4 y= sin x

F(D)≡ (D²+1)     [Here ]

Therefore

        [ replace D² by -1² and F(D)≠0]

The complete solution is

Differentiating with respect to x

The initial condition are y(0) =1 and y'(0)=4

and

Therefore the complete solution is

Case 2:

g(x)=0

Then

The complete solution is

Differentiating with respect to x

The initial condition are y(0) =1 and y'(0)=4

and

⇒B=2

Therefore the complete solution is