Since f (x, y) = 1 + y2 and "∂f/∂y" = 2y are continuous everywhere, the region r in theorem 1.2.1 can be taken to be the entire xy-plane. use the family of solutions in part (a) to find an explicit solution of the first-order initial-value problem y' = 1 + y2, y (0) = 0.

The solution to the differential equation

y' = 1 + y²

is

y = tan x

Step-by-step explanation:

Given the differential equation

y' = 1 + y²

This can be written as

dy/dx = 1 + y²

Separate the variables

dy / (1 + y²) = dx

Integrate both sides

tan^ (-1) y = x + c

y = tan (x+c)

Using the initial condition

y (0) = 0

0 = tan (0 + c)

tan c = 0

c = tan^ (-1) 0 = 0

y = tan x