1-xy=x-y^2rArr 1+y^2=x+xy=x(1+y)1−xy=x−y2⇒1+y2=x+xy=x(1+y)
rArr(1+y^2)/(1+y)=x⇒1+y21+y=x
Diff.ing both sides of this eqn. w.r.t. y , we have,
dx/dy={(1+y)*d/dy(1+y^2)-(1+y^2)*d/dy(1+y)}/(1+y)^2dxdy=(1+y)⋅ddy(1+y2)−(1+y2)⋅ddy(1+y)(1+y)2
dx/dy={(1+y)(2y)-(1+y^2)(1)}/(1+y)^2=(2y+2y^2-1-y^2)/(1+y)^2dxdy=(1+y)(2y)−(1+y2)(1)(1+y)2=2y+2y2−1−y2(1+y)2
dx/dy=(y^2+2y-1)/(1+y)^2, hence, dy/dx=(y+1)^2/(y^2+2y-1).....(star), i.e.,
dy/dx=(y^2+2y+1)/(y^2+2y-1)=(y^2+2y-1+2)/(y^2+2y-1)
=(y^2+2y-1)/(y^2+2y-1)+2/(y^2+2y-1)=1+2/((y^2+2y-1). Therefore,
(d^2y)/dx^2=d/dx(dy/dx)=d/dx{1+2/(y^2+2y-1)}
=0+2d/dx(y^2+2y-1)^-1=2{d/dy(y^2+2y-1)^-1}*dy/dx
=2{-1(y^2+2y-1)^-2*d/dy(y^2+2y-1)}dy/dx
={-2/(y^2+2y-1)^2}*2(y+1)*dy/dx
=-(4(y+1))/(y^2+2y-1)^2*(y+1)^2/(y^2+2y-1)...............[by, (star)]
:. (d^2y)/dx^2=-(4(y+1)^3)/(y^2+2y-1)^3
Hope, this will be of Help! Enjoy Maths.!