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

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Answer 1 · 2016-08-07T09:20:59

#(d^2y)/dx^2=-(4(y+1)^3)/(y^2+2y-1)^3#

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

#rArr(1+y^2)/(1+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)^2#

#dx/dy={(1+y)(2y)-(1+y^2)(1)}/(1+y)^2=(2y+2y^2-1-y^2)/(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#

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