Determine the longest interval in which the given initial value

problem is certain to have a unique twice-differentiable solution.

Do not attempt to find the solution. (Enter your answer using

interval notation.) ty” + 3y = t, y(1) = 1, y'(1) = 9

Consider the initial value problem, ty+3y-t,y(1)-1,y'(1)-9 Rewrite this differential equation as, Compare it with y+ p(t)y’+at)y- g(),y()o()-o obtained as, 08)-1 Here, p(t) and q(t) are continuous for all values of t and q(t)scontinuous except for t=0. By existence and uniqueness theorem, the longest open interval containing the initial point t 1 in which all the coefficients are continuous is, 0 < t < 0 Hence, the longest interval which is unique twice differentiable solution is(0,)