We rewrite the given eqn. as -2y=x(cosy-y), or, x=(2y)/(y-cosy)−2y=x(cosy−y),or,x=2yy−cosy.
Diff.ing w.r.t. yy both sides of this eqn., we have,
dx/dy=d/dy{(2y)/(y-cosy)}=2d/dy{y/(y-cosy)}dxdy=ddy{2yy−cosy}=2ddy{yy−cosy}.
Using Quotient Rule for Diffn. on the R.H.S., we get,
dx/dy=2[{(y-cosy)*d/dy(y)-y*d/dy(y-cosy)}/(y-cosy)^2]dxdy=2⎡⎣(y−cosy)⋅ddy(y)−y⋅ddy(y−cosy)(y−cosy)2⎤⎦
=2[{(y-cosy)(1)-(y)(1+siny)}/(y-cosy)^2]=2[(y−cosy)(1)−(y)(1+siny)(y−cosy)2]
=2{(y-cosy-y-ysiny)/(y-cosy)^2}=2{y−cosy−y−ysiny(y−cosy)2}
=-2{(cosy+ysiny)/(y-cosy)^2}=−2{cosy+ysiny(y−cosy)2}.
Therefore, dy/dx=1/(dx/dy)=-(y-cosy)^2/(2(cosy+ysiny)dydx=1dxdy=−(y−cosy)22(cosy+ysiny).
Enjoy Maths.!
