a) Verify that y = -1/(xc) is a family of one-parameter solutions of the equation differentially`=y2. b) Since f (x, y) = y2 and ∂f/∂y = 2y are continuous everywhere, the region R of Theorem 1.2.1 can be taken as the entire xy-plane. Find a solution from the family of part (a) that satisfies that y(0) = 1. Then find a solution from the family of part (a) that satisfies that y(0) = -1. Find the longest interval I of definition for the solution of each initial value problem. c) Determine the longest interval of definition I for the solution of the initial value problem y`= y2 , y(0) = 0. [Hint: The solution is not a member of the family of solutions in part (a)] .

a) To verify that y = -1/(xc) is a family of one-parameter solutions of the differential equation y' = y^2, we need to substitute y = -1/(xc) into the differential equation and see if it satisfies the equation. First, let's calculate y' by taking the derivative of y with respect to x: y' = d/dx(-1/(xc)) = (1/(x^2c)) * 1 = 1/(x^2c) Now let's substitute y = -1/(xc) and y' = 1/(x^2c) into the differential equation: y' = y^2 1/(x^2c) = (-1/(xc))^2 1/(x^2c) = 1/(x^2c^2) The equation is satisfied, so y = -1/(xc) is indeed a family of one-parameter solutions of the differential equation y' = y^2. b) To find a solution from the family y = -1/(xc) that satisfies y(0) = 1, we substitute x = 0 and y = 1 into the equation: 1 = -1/(0 * c) This equation is not defined since division by zero is not allowed. Therefore, there is no solution from the family that satisfies y(0) = 1. To find a solution from the family y = -1/(xc) that satisfies y(0) = -1, we substitute x = 0 and y = -1 into the equation: -1 = -1/(0 * c) This equation is also not defined since division by zero is not allowed. Therefore, there is no solution from the family that satisfies y(0) = -1. Both initial value problems do not have solutions from the given family. c) The initial value problem y' = y^2, y(0) = 0 is a special case that is not a member of the family y = -1/(xc). To solve this initial value problem, we can separate variables and integrate: ∫1/y^2 dy = ∫1 dx -1/y = x + C y = -1/(x + C) To find the longest interval of definition I for this solution, we need to consider the denominator of the expression -1/(x + C). The denominator cannot be zero, so x + C ≠ 0. Therefore, the longest interval of definition I is (-∞, ∞), which covers the entire real number line. So, the longest interval of definition for the solution of the initial value problem y' = y^2, y(0) = 0 is (-∞, ∞).