Exact Equations — Question 8

Question 8 – Solution

Strategy. Combine the logarithms using the product $u=xy$, but keep the connected domain and the sign of $u$ in view.

Step 1: Construct the potential. Both cross partials are $1$. Integration gives $$\boxed{F=\ln|x|+\ln|y|+xy=\ln|xy|+xy}.$$ Indeed $F_x=1/x+y$ and $F_y=1/y+x$ wherever $xy\ne 0$.

Step 2: Select the positive branch. The data $(1,2)$ give $F=\ln 2+2$. For $u>0$, the function $\phi(u)=\ln u+u$ is strictly increasing, since $\phi'(u)=1/u+1>0$. Thus $\phi(u)=\phi(2)$ forces $u=2$. Consequently $$\boxed{y=2/x,\qquad I=(0,\infty)}.$$ Substitution gives $M=3/x$, $N=3x/2$ and $y'=-2/x^2$, so $M+Ny'=0$. The branch cannot cross $x=0$: the equation is undefined there and $y\to+\infty$ from the right. The negative-quadrant branch $y=2/x$, $x<0$, is disconnected from the initial point.

Step 3: Analyze the degenerate negative level. For $(1,-1)$, $F=-1$. On $u<0$, $\phi(u)=\ln(-u)+u$ has derivative $1/u+1$, positive for $u<-1$ and negative for $-1<u<0$. Its unique maximum is $\phi(-1)=-1$. Therefore this level forces $xy=-1$, giving $$\boxed{y=-1/x\quad(x>0)}.$$ Along this graph both $M$ and $N$ are zero, so the original equation is satisfied directly. Every solution through the point must conserve $F$ and remain in the same quadrant, which proves this branch is unique there despite the vanishing coefficients.