Let y = y(x) be the solution of the differential equation dy dx = 2 x (1 + y^2) with the initial condition y(0) = 0 . If (-a, a) is the maximal open interval on which the solution y(x) is defined, then the value of a^2 is equal to:

Step-by-step solution

The given differential equation is dy dx = 2 x (1 + y^2) . Separating the variables: dy 1 + y^2 = 2 x dx Integrating both sides: dy 1 + y^2 = 2 x dx ⁻¹(y) = 4 x^2 + C Using the initial condition y(0) = 0 : ⁻¹(0) = 0 + C C = 0 Thus, the solution is given by: ⁻¹(y) = 4 x^2 For the function y(x) = ( 4 x^2 ) to be defined and continuous, the argument of the tangent function must strictly lie in the principal range of the inverse tangent function, which is (- 2 , 2 ) . Therefore, we must have: - 2 Since x^2 0 , the left