On the front of your bluebook write: (1) your name, (2) your student id number, (3) lecture section, (4) your instructor"s name, and (5) a grading table for eight questions. Text books, class notes, and calculators are not permitted. The initial condition y = ey(1 + et) Therefore p(t) = 2t/(t2 1) and the integrating factor is. (t) = expz p(t)dt = exp ln(t2 1) = (t2 1) Thus giving general solution ( (t)y(t)) = t y(t)) = 2 t2 + c t2 1 s. t. y(t)) = Solution: (a) note that y1 and y4 satisfy the homogeneous equation t2y 2ty + 2y = 0. y2 and y3 do not. Therefore, the homogeneous solution is given by yh(t) = c1t + c2t2. (b) we rst note that variation of parameters requires a leading coef cient of 1. Therefore we divide both sides of the differential equation by t2 to obtain.
