Let y = y(x) be the solution of the differential equation dy dx + 2xy = 4x^3 satisfying y(0) = 1 . Then the minimum value of the function y(x) is

Step-by-step solution

Given differential equation is dy dx + 2xy = 4x^3 . This is a linear differential equation of the form dy dx + P(x)y = Q(x) , where P(x) = 2x . Integrating Factor (I.F.) = e^ 2x dx = e^ x^2 The solution is given by: y e^ x^2 = 4x^3 e^ x^2 dx + C To evaluate the integral, let x^2 = t 2x dx = dt . The integral becomes: 2t e^t dt = 2(t e^t - e^t) = 2(x^2 - 1)e^ x^2 So, y e^ x^2 = 2(x^2 - 1)e^ x^2 + C y = 2(x^2 - 1) + C e^ -x^2 Using the initial condition y(0) = 1 : 1 = 2(0 - 1) + C C = 3 The particular solution is: y(