How do you find $(d^2y)/(dx^2)$ given $x+siny=xy$ ?

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Answer 1 · 2016-09-05T19:00:36

$(d^2y)/dx^2=[(y-1){(y-1)siny+2(cosy-x)}]/(cosy-x)^3$ .

$x+siny=xy rArr siny=xy-x=x(y-1) rArr x=siny/(y-1)$

$rArr d/dy(x)=d/dy(siny/(y-1))$

$rArr dx/dy={(y-1)d/dy(siny)-sinyd/dy(y-1)}/(y-1)^2$

$={(y-1)cosy-siny}/(y-1)^2$

Since, $siny=x(y-1)$ , we have,

$dx/dy={(y-1)cosy-x(y-1)}/(y-1)^2=(cosy-x)/(y-1)$

$:. dy/dx=(y-1)/(cosy-x).......................................................(star)$

$rArr y'(cosy-x)=y-1$

$:. d/dx{y'(cosy-x)}=d/dx(y-1)$

$:. y'd/dx(cosy-x)+(cosy-x)d/dx(y')=dy/dx-0=y'$ .

$:. y'{d/dxcos y-d/dx(x)}+(cosy-x)y''=y'$ .

$:. y'{(-siny)y'-1}+(cosy-x)y''=y'$ .

$:. (cosy-x)y''=y'(y'siny+1)+y'=y'(y'siny+2)$ .

$=(y-1)/(cosy-x){((y-1)siny)/(cosy-x)+2}.....................[by (star)]$

$:. y''=(y-1)/(cosy-x)^2{((y-1)siny)/(cosy-x)+2}$

$rArr (d^2y)/dx^2=y''=[(y-1){(y-1)siny+2(cosy-x)}]/(cosy-x)^3$ .

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How do you find (d^2y)/(dx^2)(d^2y)/(dx^2) given x+siny=xyx+siny=xy?