Find $y''$ in terms of $x$ and $y$
Find $y''$ in terms of $x$ and $y$
$$xy=x+y$$
I have calculated $y'$ as
$$\frac{y-1}{1-x}=y'$$
which is correct according to the solutions manual but when I try to calculate $y''$ I get
$$y''= \frac{y'(1-x)+y-1}{(1-x)^2}$$
$$y''= \frac{y'+y-1}{1-x}$$
but the solutions manual get $y''= \frac{2y'}{1-x}$
What am I doing wrong?
$\endgroup$ 13 Answers
$\begingroup$Do not solve for $y'$ that only makes matters more complicated. Just differentiate twice and solve for $y''$.
$$xy=x+y \implies y+xy'=1+y' \implies y'+y'+xy''=y''$$
Having these equations you can then solve the separatly for the derivatives:
$$y'=\frac{y-1}{1-x}$$ $$y''=\frac{2}{1-x}y'=\frac{2}{1-x}\frac{y-1}{1-x}$$
$\endgroup$ 1 $\begingroup$Note that $$\frac {y'(1-x)+y-1}{(1-x)^2}=\frac {y'}{1-x}+\frac {y-1}{(1-x)^2}=2\frac {y'}{1-x}$$ because of the formula you already have for $y'$. So the two answers are the same.
$\endgroup$ $\begingroup$Up to $$ y''= \frac{y'(1-x)+y-1}{(1-x)^2} $$ it looks good. However, from there you simplify the fraction by $1-x$ without taking the $y-1$ in the numerator into account. So instead of $y'' = \frac{y' +y-1}{1-x}$ that you have, it should've been $$ y''= \frac{y'+\frac{y-1}{1-x}}{1-x} $$ and finally, we use that we already know that $y' = \frac{y-1}{1-x}$ to make the numerator into $2y'$.
$\endgroup$
