Obtain an explicit function from an implicit expression

Question

Let $y=y(x)$ be a function implicitly defined as $$xy+\ln(xy)=1$$ near a point $P(1,1)$.

I have to find the explicit expression $y(x)$ as well as the values $y'(1)$ and $dy(1)$.

I've tried applying the exponential to both sides but I could not find a solution: $$ e^{xy}xy=e $$ doesn't seem to be very helpful.

Also, using implicit differentiation didn't help me that much with obtaining $y'(1)$. If I am not mistaking, by implicit differentiation one obtains $$y+xy'+\frac{1}{x}+\frac{y'}{y}=0$$

But I still can't see how I can obtain the value $y'(1)$ from this. As for $dy(1)$: I have no idea how I could obtain that!

Any help/suggestions would be appreciated.

Answer 1

it must be $$y(x)+xy'(x)+\frac{1}{xy(x)}\cdot (y(x)+xy'(x))=0$$ comment: your second equation must be $$e^{xy}=e^{1-\ln(xy)}$$ the equation $$e^{xy}xy=e$$ is right now.

Answer 2

Hint You're on the right track to find $y$: We can rewrite the left-hand side of the expression $$y + x y' + \frac{1}{x} + \frac{y'}{y} = 0$$ you produce when implicitly differentiation the original formula as $$\left(1 + \frac{1}{xy}\right)(y + x y') = 0 .$$ There is a one-parameter family of solutions $y(x)$ that satisfy this (ordinary differential) equation in $y$, but since we discarded some information when differentiating, not all of these solutions need satisfy the original equation. So, we must substitution our general solution into the original equation and see for which parameter values that equation is satisfied (it turns out that it does for precisely one).

Answer 3

So far you have $$y+xy'+\frac{1}{x}+\frac{y'}{y}=0\\y'(1+xy)+y^2+\frac yx=0\\y'(1+xy)+\frac {y(xy+1)}x=0\\(xy'+y)(1+xy)=0\\y=-\frac1x\text{ or } (xy)'=0\implies xy=c\implies y=\frac cx$$So the general solution is $y=\frac cx$.

Substituting back into the original equation, $$xy+\ln(xy)=c+\ln c=1$$The only solution to this is $c=1$. This can be seen by comparing the graphs of $\ln x$ and $1-x$ - they only intersect in one place, and that is at $x=1$.

Answer 4

The solution to $$u+\ln(u)=1$$ or equivalent $$ue^u=e$$ is given by the Lambert-W function, $$u=W_0(e)=1.$$ As for positive arguments there is only one real branch of that function, the solution is unique and $$ xy(x)=1\iff y(x)=\frac{1}{x}. $$