[xy=log y+C],[y-cos y=x]quad :quad y'=(y^(2))/(1-xy)quad (xy!=1)=

xy = log y + C : y' = (y^(2))/(1 - xy)(xy ne 1)

xy = log y + C : y' = (y^(2))/(1 - xy)(xy ne 1)

xy = log y + C : y' = (y^(2))/(1 - xy)(xy ne 1)

xy = log y + C : y' = (y^(2))/(1 - xy)(xy ne 1)

Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: xy=log y+Cvdotsy'=(y^(2))/(1-xy)(xy!=1)

verify that the given functions(explicit or implicit) is a solution of the corresponding differential equation : xy = log y + C : y' = (y^(2))/(1 - xy)(xy ne 1)

Verify that the given function (explicit or implicit) is a solution of the correseponding differential equation : xy = log y + c : y'= y^2/(1-xy) (xy ne 1)

If xy-log y=1,theny^(2)+(xy-1)(dy)/(dx)=

log(xy)=x^(2)-y^(2) when x=1,y=1

If xy log(x+y)=1, prove that (dy)/(dx)=-(y(x^(2)y+x+y))/(x(xy^(2)+x+y))