How do you determine (d^y)/(dx^2)dydx2 given 1-xy=x-y^21xy=xy2?

1 Answer
Aug 7, 2016

(d^2y)/dx^2=-(4(y+1)^3)/(y^2+2y-1)^3d2ydx2=4(y+1)3(y2+2y1)3

Explanation:

1-xy=x-y^2rArr 1+y^2=x+xy=x(1+y)1xy=xy21+y2=x+xy=x(1+y)

rArr(1+y^2)/(1+y)=x1+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+2y21y2(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.!