Find the particular solutions of the …

dx/dy=tan^-1y-x/1+y^2 =tan^-1y/1+y^2 - x/1+y^2 dx/dy + x/1+y^2 = tan^-1y/1+y^2 This is a linear differential equation of the form dx/dy+Px=Q Where P=1/1+y^2 and Q=tan^-1y/1+y^2 Integrating factor(I.F)=e^integral of 1/1+y^2 dy =e^tan^-1y Therfore the solution of this differential equation is given by, x(I.F)=integral of Q ×(I.F) dy +C x.tan^-1y=integral of tan^-1y/1+y^2 × e^tan^-1y dy +C put tan^-1y=t And, 1/1+y^2dy=dt x e^t=integral of t × e^t dt +C On RHS side use by part rule and we get x e^t= e^t(t-1) +C On division by e^t both side we get x=(t-1) +C.e^-t Put t=tan^-1y x=( tan^-1y -1) + C.e^-tan^-1y This is the required solution.