find the general solution of the given differential equation
1. 2y''+3y'+y=t^2 +3sint
find the solution of the given initial value problem
1. y''−2y'−3y=3te^2t, y(0) =1, y'(0) =0
2. y''−2y'+y=te^t +4, y(0) =1, y'(0) =1
Again considering y'' + 4y' + 3y = 0:
(a) Solve the IVP y'' + 4y' + 3y = 0; y(0) = 1, y'(0) = α where
α > 0.
(b) Determine the coordinates (tm,ym) of the maximum point of
the solution as a function of α.
(c) Determine the behavior of tm and ym as α →∞.