Question

Find y' and y" where y - ln y = xy.

Answer 1

For y':

\[-lny=y(x-1)\]\[\frac{-lny}{y}=x-1\] \[(-ln{y})y^{-1}=x-1\]

Now use the quotient rule for the LHS

\[\frac{(lny-1)}{y^2}=1\] \[(lny-1)=y^2\] \[lny-y^2=1\]

For y'':

\[lny-y^2=1\]

\[\frac{1}{y}-2y=0\]\[\frac{1}{y}=2y\]\[1=2y^2\]\[1/2=y^2\]\[\pm \sqrt{\frac{1}{2}}=y\]