Exact Equations — Question 2

Question 2 – Solution

Strategy. A partial antiderivative permits an arbitrary function of the other variable; matching the second coefficient determines it.

Step 1: Locate the missing term. Both cross partials equal $e^y+2y$, so the form is exact on $\mathbb R^2$. Integration with respect to $x$ must give $$F=xe^y+xy^2+h(y).$$ Then $F_y=xe^y+2xy+h'(y)$, so $h'(y)=\cos y$. We may take $h(y)=\sin y$; treating it as merely a numerical constant would omit a required derivative.

Step 2: Apply and test the initial condition. The correct potential is $F=xe^y+xy^2+\sin y$, and $F(0,0)=0$. Thus $$\boxed{x(e^y+y^2)+\sin y=0}.$$ Its derivative along a graph is exactly the given equation. At the initial point, $M=1$ and $N=1$, so the local graph exists and $\boxed{y'(0)=-1}$.

In contrast, $F_*=0$ says $x(e^y+y^2)=0$. Since $e^y+y^2>0$, this is the vertical line $x=0$, which cannot define $y(x)$ on an open interval. Also $(F_*)_y=N-\cos y$, so its differential is not the stated form.

Step 3: Check by reversing the integration order. Integrating $N$ with respect to $y$ gives $$F=xe^y+xy^2+\sin y+g(x).$$ Now $F_x=e^y+y^2+g'(x)=M$ requires $g'(x)=0$, leaving only an additive constant. Any two potentials for the same form have difference with both partial derivatives zero; on the connected plane that difference is constant. Changing this constant only relabels the same solution levels.