Take the Laplace transform of the following initial value and solve for Y(s)=L{y(t)}: sin(π t), 0 ≤ t <1
y''+9y= { y(0)=0, y′(0)=0
0, 1 ≤ t
a) Y(s)= ? (Hint: write the right hand side in terms of the Heaviside function)
b) Now find the inverse transform to find y(t)= ? . (Use step(t-c) for uc(t) .)
Note= π/ (s^2+π^2)(s^2+9) = π/ π^2 -9 (1/(s^2+9) - 1/(s^2+ π^2)