In problems 1 to 7, solve all initial value problems. For each problem, state the largest interval in which the solution function is defined. 1. y"+y'= 3e(t/2) y(0) = 4 y'(0) = 3 2. ty"- y=t +t y(1) = 1 y'(1) = 5 3. y" - y' - 2y = 0 y(0) = 2 y'(0) = 7 4. (12 + 9) y" + 2ty' = 0 y(3) = 2n y'(3) = 2/3 5. y" + y' + 12y = 0 y(IT) = -2 y'(n) = -1 6. y" + 2y' - 3y = 0 y(27) = 1 y'(2nt) = 1 7. y" + 2y' - 4y = 0 y(2) = 1 y'(0) = -1