Exact Equations — Question 4

Question 4 – Solution

Strategy. Use exactness as an equation for the missing coefficient, then distinguish a complete level curve from its graph branches.

Step 1: Recover the form. Exactness gives $N_x=M_y=1$, so $N=x+h(y)$. The data force $h(y)=2y$, uniquely. A potential is $$\boxed{N=x+2y,\qquad F=x^2+xy+y^2}.$$ Its derivatives are $2x+y$ and $x+2y$, as required.

Step 2: Select a branch. The initial point gives $F=1$. Solving the quadratic in $y$ and selecting $y(0)=1$ yields $$\boxed{y=\frac{-x+\sqrt{4-3x^2}}2,\qquad I=(-2/\sqrt 3,2/\sqrt 3)}.$$ At $(0,1)$, $y'=-(2x+y)/(x+2y)=-1/2$.

Step 3: Interpret the endpoints and full curve. On this branch, $N=\sqrt{4-3x^2}>0$. At the two endpoints, $N=0$, while $M=3x/2\ne 0$. Thus a finite $y'$ cannot satisfy $M+Ny'=0$ there. The full ellipse has vertical tangents at $$\boxed{(2/\sqrt 3,-1/\sqrt 3),\quad(-2/\sqrt 3,1/\sqrt 3)}.$$ It remains a smooth curve at these points, but cannot be continued through either as a differentiable graph over $x$. Most vertical lines through the ellipse meet both quadratic branches, so the complete ellipse is not one function $y(x)$.