Consider y y deltat pi2 alpha u pi2t y0 0 y0 1 Solve t

Consider y\" + y = delta(t - pi/2) + alpha u pi/2(t), y(0) = 0, y\'(0) = 1. Solve the initial value problem. Find the value of alpha for which y(3pi/2) = 1.

Solution

take laplace transformation

s^2Y(s) -s*0 - 1 + Y(s) = e^(-pi/2 s) + alpha e^(-pi/2 s) /s

Y(s) (s^2+1) = e^(-pi/2 s) (1+alpha/s)

Y(s) = e^(-pi/2 s)*(1+alpha/s)/(s^2+1)

Y(s) = e^(-pi/2 s)*[1/1+s^2 + alpha/s -alpha*s/(1+s^2)]

take inverse transformation

y(t) = u_pi/2(t) [sint + alpha - alpha cost]

value of alpha for y(3pi/2) = 1 is 2