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3. Even if we are unable to calculate a messy integration, we may still be able to express the solutions to differential equations as integrals. The idea is to take advantage of the Fundamental Theorem of Calculus: f f (:17) d9: = f: f (t) dt Where we are free to choose the value of a arbitrarily. (a) Verify that the function 1' 1 y(a:) = e%/ e_%dt 1 t is the solution to the initial value problem w\".u\"+rv+iv2 = (1+2w)y, 9(1) =0 (b) Find the solution to the initial value problem, do: + cos(ac)dy = y cos(m)dm , y(1r) = 2