Let y = y(x) be the solution of the differential equation x(x^2+1) dy dx + 2y = 3x(x^2+1)^2 for x > 0 . If y(1) = 4 , then the area bounded by the curve y = y(x) , the x-axis, and the vertical lines x=1 and x=2 is

Step-by-step solution

The given differential equation is: x(x^2+1) dy dx + 2y = 3x(x^2+1)^2 dy dx + 2 x(x^2+1) y = 3(x^2+1) This is a linear differential equation of the form dy dx + P(x)y = Q(x) . Integrating Factor (I.F.) = e^ 2 x(x^2+1) dx = e^ 2x x^2(x^2+1) dx Let x^2 = t 2x dx = dt I.F. = e^ 1 t(t+1) dt = e^ ( 1 t - 1 t+1 ) dt = e^ ( t t+1 ) = x^2 x^2+1 The solution is given by: y ( I.F. ) = Q(x) ( I.F. ) dx y ( x^2 x^2+1 ) = 3(x^2+1) x^2 x^2+1 dx y ( x^2 x^2+1 ) = 3x^2 dx = x^3 + C Given y(1) = 4 : 4 ( 1 2 ) = 1 + C C = 1 So, the