Question

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

Answer 1

Explanation:

`g(x)=y''+4y\\\\y''+4y=0\\\\m^2+4=0\\m_(1,2) =+2i,-2i`

so the complimentary solution is

`y_(c) =c_(1) cos2x+c_(2) sin2x`

solving for particular solution we get

`y_(p) =acosx+bsinx`

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

`-acosx-bsinx+4acosx+4bsinx=sinx\\`

as g(x)=sin x

so above equation yields us

`a=0,b=1/3`

so writing the solution

`y=y_(c) +y_(p) \\y=c_(1) cos2x+c_(2) sin2x+(1/3)sinx`

using IC's to get the values for
`c_(1) ,c_(2)`

`c_(1)=1 ,c_(2) =5/6\\so\\y(x)=cos2x+(5/6)sin2x+(1/3)sinx ,0\leq x\leq (\pi )/(2)`

solving for upper range

`y(x)=(2)/(3) cos2x+(5)/(6) sin2x,x>(\pi )/(2)`