Let y = y(x) be the solution to the differential equation dy dx + 2x^2 - 1 x(x^2 - 1) y = x^2 x^2 - 1 for x > 1 . If _ x 1^+ x-1 y(x) = 0 , then the local minimum value of the function y(x) for x > 1 is

Step-by-step solution

The given differential equation is linear of the form dy dx + P(x)y = Q(x) . Here, P(x) = 2x^2 - 1 x(x^2 - 1) = x^2 + (x^2 - 1) x(x^2 - 1) = x x^2 - 1 + 1 x . The integrating factor (IF) is: IF = e^ P(x) dx = e^ ( x x^2 - 1 + 1 x ) dx IF = e^ 1 2 (x^2 - 1) + x = e^ (x x^2 - 1 ) = x x^2 - 1 . The general solution is given by: y x x^2 - 1 = x^2 x^2 - 1 x x^2 - 1 dx + C y x x^2 - 1 = x^3 x^2 - 1 dx + C To evaluate the integral, let x^2 - 1 = t 2x dx = dt x dx = dt 2 . x^2 x^2 - 1 x dx = t + 1 t dt 2 = 1 2 ( t^ 1/2 + t