Let a curve y=y(x) pass through the points ( 2 , 1 ) and ( , 2) . If y(x) satisfies the differential equation (y^2+1) dy dx = x 1+ x (2+ 1+ x )^2 where is a constant, then the value of is equal to :

Step-by-step solution

Separating the variables, we get: (y^2+1) dy = x 1+ x (2+ 1+ x )^2 dx Integrating both sides: (y^2+1) dy = x 1+ x (2+ 1+ x )^2 dx For the right hand side, let t = 2+ 1+ x . Differentiating with respect to x : dt = 1 2 1+ x (- x) dx x 1+ x dx = -2 dt Substituting this into the integral: y^3 3 + y = t^2 (-2) dt y^3 3 + y = -2 ( - 1 t ) + C y^3 3 + y = 2 2+ 1+ x + C Given that the curve passes through ( 2 , 1 ) : 1 3 + 1 = 2 2+ 1+0 + C 4 3 = 2 3 + C (1) Given that the curve passes through ( , 2) : 8 3 + 2 = 2 2+ 1-1 +