The particular solution of differential equation (1+y^2 )(1+ x) d x+x ~d y=0 at x=1, y=1 is

Step-by-step solution

aligned & (1+y^2 )(1+ x) d x+x ~d y=0 & (1+y^2 )(1+ x) d x=-x ~d y & ( 1+ x x ) d x= ( -1 1+y^2 ) d y aligned Integrating on both sides, we get (1+ x) x ~d x=-1 1 1+y^2 ~d y tdt =- ⁻¹ y+ c [ array l Let 1+ x= t 1 x ~d x= dt array ] aligned & t ^2 2 =- ⁻¹ y+ c ...(i) & (1+ x)^2 2 =- ⁻¹ y+ c aligned aligned & At x=1, y=1 & (1+ 1)^2 2 =- ⁻¹(1)+c & c= 1 2 + 4 aligned Substituting above value in (i), we get aligned & (1+ x)^2 2 =- ⁻¹ y+ 1 2 + 4 & 1 2 + x+ ( x)^2 2 =- ⁻¹ y+ 1 2 + 4 & x+ ( x)^2 2 + ⁻¹ y= 4 aligned