Consider the initial value problem ty 3y cost4 with 1 Give

Consider the initial value problem ty\' + 3y = cos(t)4 with =1 Give an interval on which the initial value problem has a unique solution

Solution

ty\' + 3y = tCos(t^4)

y\' + (3/t)y = Cos(t^4)

3/t is discontinuous at t = 0 and Cos(t^4) is always continuous function.

Hence, the differential equation is valid for -<t<0 and 0<t<.

Since, ()^1/4 lies in 0<t<, hence there is a unique solution to the differential equation in 0<t<.