A designer draws curves of a chart on which the equation y² dx + (x² - xy) dy = 0 holds, with x > 0…

Step 1. The factor. $Mx + Ny = xy^2 + x^2y - xy^2 = x^2y$. So $\mu = \dfrac{1}{x^2y}$.

Step 2. The new equation. $\dfrac{y}{x^2}\,dx + \left(\dfrac{1}{y} - \dfrac{1}{x}\right)dy = 0$. Now $M_y = \dfrac{1}{x^2}$ and $N_x = \dfrac{1}{x^2}$, so the equation is exact.

Step 3. The solution. $F = -\dfrac{y}{x} + g(y)$ and $F_y = -\dfrac{1}{x} + g' = \dfrac{1}{y} - \dfrac{1}{x}$. So $g = \ln y$ and $\ln y - \dfrac{y}{x} = C$.

Step 4. The constant. At $(1, 1)$: $C = 0 - 1 = -1$.

Step 5. Solve for $x$. From $\ln y - \dfrac{y}{x} = -1$ we get $x = \dfrac{y}{1 + \ln y}$. At $y = 2$: $x = \dfrac{2}{1.693147} = 1.1812$. At $y = 0.5$: $x = \dfrac{0.5}{1 - 0.693147} = 1.6294$.