Let y = y(x) be the solution of the differential equation (x^2+1) dy dx - 2xy + 2x^5 + 4x^3 + 2x = 0 satisfying y(0) = 4 . The local maximum value of the function y(x) is

Step-by-step solution

The given differential equation is: (x^2+1) dy dx - 2xy + 2x(x^4 + 2x^2 + 1) = 0 (x^2+1) dy dx - 2xy = -2x(x^2+1)^2 Dividing by x^2+1 , we get: dy dx - 2x x^2+1 y = -2x(x^2+1) This is a linear differential equation with P(x) = - 2x x^2+1 . IF = e^ - 2x x^2+1 dx = e^ - (x^2+1) = 1 x^2+1 The general solution is: y 1 x^2+1 = -2x(x^2+1) 1 x^2+1 dx y x^2+1 = -2x dx = -x^2 + C Using the initial condition y(0) = 4 : 4 0+1 = -0 + C C = 4 Thus, the function is: y x^2+1 = 4 - x^2 y = (4-x^2)(x^2+1) = -x^4 + 3x^2 + 4 To find