Let x = x(y) be the solution of the differential equation (1+y^2) dx = (2yx + 4y^3) dy with the initial condition x = 2 when y = 0 . Then the value of x when y = e-1 is equal to

Step-by-step solution

The given differential equation can be rewritten as: dx dy - 2y 1+y^2 x = 4y^3 1+y^2 This is a linear differential equation of the form dx dy + P(y)x = Q(y) , where P(y) = - 2y 1+y^2 and Q(y) = 4y^3 1+y^2 . The integrating factor (IF) is: IF = e^ P(y) dy = e^ - 2y 1+y^2 dy = e^ - (1+y^2) = 1 1+y^2 The general solution is given by: x IF = Q(y) IF dy x 1+y^2 = 4y^3 (1+y^2)^2 dy Let 1+y^2 = u 2y dy = du . Also, y^2 = u-1 . 2(u-1) u^2 du = ( 2 u - 2 u^2 ) du = 2 |u| + 2 u + C Substituting back u = 1+y^2 : x 1+y^2 = 2 (