Exact Equations — Question 5

Question 5 – Solution

Strategy. The selected level has crossing branches. Analyze differentiability at their intersection directly in the original equation.

Step 1: Find the level. Since $M_y=N_x=0$, a potential is $F=y^2-x^2$. Along any solution, $dF/dx=0$, and the initial condition gives $$\boxed{y^2-x^2=0}.$$ Thus for each $x\ne 0$, $y/x$ is either $1$ or $-1$.

Step 2: Classify differentiable choices. Continuity makes $y/x$ constant on each of the two connected pieces $x<0$ and $x>0$: a continuous function taking only the values $\pm 1$ cannot switch between them. Write those constants as $s_-$ and $s_+$. At $0$, the left and right difference quotients are $s_-$ and $s_+$. Differentiability requires them to agree. Therefore the only possibilities are $$\boxed{y=x\quad\text{or}\quad y=-x}.$$ Both extend to all of $\mathbb R$.

Step 3: Verify and identify the failures. For either line, $yy'=x$ everywhere, including $0$. The choices $|x|$ and $-|x|$ satisfy the level identity and the equation away from $0$, but their one-sided derivatives disagree at $0$. They fail the required definition.

Step 4: Explain the degenerate point. Dividing gives $y'=x/y$, which is undefined at the initial point and discards information about solutions of the original form there. Also $F_y=2y=0$ at $(0,0)$, so the nonzero-$F_y$ criterion does not apply. Its failure does not prove nonexistence: here two valid graph branches pass through the point.