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) = 6, where g(x) = sin(x), 0 ≤ x ≤ π/2 0, x > π/2

Step-by-step explanation:

so the complimentary solution is

solving for particular solution we get

to find the values of constants a,b we put values in main equation and

as g(x)=sin x

so above equation yields us

so writing the solution

using IC's to get the values for

solving for upper range

answer: i'm going in

step-by-step explanation:

x= -3

step-by-step explanation:

step 1: simplify brackets

4 - x - 1 = 6

step 2: simpilfy

3 - x = 6

step 3: subtract 3 from both sides

-x = 6 - 3

step 4: simpilfy 6 - 3

-x = 3

step 5: multiply both sides by -1

x = -3