If the general solution of the differential equation y^ = y x + ( x y ) , for some function , is given by y |c x|=x , where c is an arbitrary constant, then (2) is equal to:

Step-by-step solution

Given d y d x = y x + ( y x ) Let ( y x )=v so that y=x v or d y d x =x d v d x +v from (1) & (2), x d v d x +v=v+ ( 1 v ) or, d v ( 1 v ) = d x x Integrating both sides, we get d x x = d v ( 1 v ) x+c= d v ( 1 v ) (where c being constant of integration) But, given y = x |c x| is the general solution so that x y = 1 v = |c x|= d v ( 1 v ) Differentiating w.r.t v both sides, we get ( 1 v )= -1 v^2 ( x y )=- y^2 x^2 when x y =2 i.e. (2) =- ( y x )^2=- ( 1 2 )^2= ( -1 4 )