Let the solution curve y=f(x) of the differential equation d y d x - y x = 2 + cosec x (1 + 2 cosec x)^2 , x (0, 2 ] pass through the point ( 2 , 0 ) . Then the area bounded by the curve y=f(x) , the x -axis, and the lines x= 6 and x= 2 is equal to :

Step-by-step solution

The given differential equation is a linear differential equation of the form d y d x + P(x)y = Q(x) . Here, P(x) = - x . Integrating Factor (IF) = e ^ - x d x = e ^ - ( x) = cosec x . Multiplying the differential equation by the IF, we get: y cosec x = 2 + cosec x (1 + 2 cosec x)^2 cosec x d x Converting to sine, we have: y cosec x = 2 x + 1 ( x + 2)^2 d x Observe that d d x ( - x x + 2 ) = -(- x)( x + 2) - (- x)( x) ( x + 2)^2 = ^2 x + 2 x + ^2 x ( x + 2)^2 = 2 x + 1 ( x + 2)^2 . Thus, y cosec x = - x x + 2 + C .