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

Step-by-step 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 $c_{1} ,c_{2}$

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

solving for upper range

$y(x)=\frac{2}{3} cos2x+\frac{5}{6} sin2x,x>\frac{\pi }{2}$