Exact Equations — Question 7

Question 7 – Solution

Strategy. Separate two different questions: whether the displayed coefficients are partial derivatives of a potential, and whether multiplying the equation changes its solutions.

Step 1: Test each form. For (A), $M_y=N_x=0$, with potential $$F=\frac{x^2+y^2}{2}.$$ For (B), $M_y=0$ and $N_x=e^x y$, which are not equal on the plane, so (B) is not exact there. For (C), both cross partials are $2xy$, and a potential is $$G=\frac{x^2+y^2}{2}+\frac{(x^2+y^2)^2}{4}.$$ Differentiating $G$ gives the two coefficients in (C).

Step 2: Establish equivalence. The equations for a graph are respectively $$x+yy'=0,\qquad e^x(x+yy')=0,\qquad (1+x^2+y^2)(x+yy')=0.$$ Both multiplying factors are strictly positive at every real point. Each equation therefore holds exactly when the first does. Exactness of a particular form is not required for it to share the same solutions as an exact form.

Step 3: Select the common IVP solution. The first potential gives $x^2+y^2=4$. Continuity from $y(0)=2$ selects $$\boxed{y=\sqrt{4-x^2},\qquad I=(-2,2)}.$$ Here $y'=-x/\sqrt{4-x^2}$ verifies all three equations. At either endpoint, $y=0$ but $x\ne 0$, so no finite derivative can satisfy $x+yy'=0$; a differentiable graph cannot extend through it.

Step 4: Generalize using the chain rule. Choose an antiderivative $H$ of the continuous function $h$. Then $$d(H(F))=H'(F)\,dF=h(F)(M\,dx+N\,dy).$$ This gives a potential for the new form. If $h(F)$ is nonzero throughout the region considered, division is valid there and the graph equations are equivalent. Allowing a zero factor can introduce extra solutions, so equivalence cannot be concluded from exactness alone.