(a) what can you say about a solution of the equation y' = −(1/5)y2 just by looking at the differential equation? the function y must be equal to 0 on any interval on which it is defined. the function y must be strictly increasing on any interval on which it is defined. the function y must be increasing (or equal to 0) on any interval on which it is defined. the function y must be strictly decreasing on any interval on which it is defined. the function y must be decreasing (or equal to 0) on any interval on which it is defined. (b) verify that all members of the family y = 5/(x + c) are solutions of the equation in part (a). y = 5 x + c ⇒ y' = − (x + c)2 . lhs = y' = − (x + c)2 = − 1 5 x + c 2 = − 1 5 y2 = rhs (c) can you think of a solution of the differential equation y' = −(1/5)y2 that is not a member of the family in part (b)? y = 0 is a solution of y' = −(1/5)y2 that is not a member of the family in part (b). y = x is a solution of y' = −(1/5)y2 that is not a member of the family in part (b). y = 5 is a solution of y' = −(1/5)y2 that is not a member of the family in part (b). every solution of y' = −(1/5)y2 is a member of the family in part (b). y = e5x is a solution of y' = −(1/5)y2 that is not a member of the family in part (b). (d) find a solution of the initial-value problem. y' = −(1/5)y2 y(0) = 0.2 y =