1. The problem statement, all variables and given/known data determine the largest interval on which a unique solution exists: y' + (t/((t^2)+1)) y = sin t with initial conditions: y(-2) = 1 2. Relevant equations arent there discontinuities at t = 0 and t = -1? (textbook says no, based on answer..i am confused why not?) i think it may have something to do with not understanding how the initial conditions, eg, y(-2) = 1 relate to the interval. 3. The attempt at a solution the equation is already in the form y' + p(t)y = g(t) p(t) = (( t/(t^2 ))+1)) if there are discont @ t = 0 and t= -1, then the intervals would be: (- infin, -1) (-1, 0) (0, infin) therefore largest interval for t = -2 is: - infin < t < -1 the answer in back though says largest interval is: - infin < t < infin please help, thanks alot. EDIT: i glossed over the square of t, so i now know that t= -1 does not make p(t) = 0.. but i am still unclear as to why t = 0 is not a discontinuity.. Also, i think i am unclear about what precisely is a discontinuity. thanks again for any help..