Question bank — Exact equations — exactness condition, finding potential function

$Mdx+Ndy=0$ is exact if $M$ and $N$ are both continuous. False ::: Continuity alone is not enough; exactness needs $M_y=N_x$ so that $M,N$ are the two partials of one potential $F$. Continuity only lets us state the test.

$M_y=N_x$ everywhere on the whole plane, then a single-valued potential $F$ definitely exists. True ::: The plane is simply-connected (no holes), so the curl-free condition $M_y=N_x$ is not just necessary but also sufficient — see Conservative Vector Fields & Potential Functions.

$Mdx+Ndy=0$ depends on how you scale the equation (e.g. multiplying both sides by $x$). True ::: Multiplying by a function changes $M$ and $N$, so it changes $M_y$ and $N_x$. A non-exact equation can become exact (that scaling factor is an integrating factor), and an exact one can be spoiled.

$Mdx+Ndy=0$, is already exact. True ::: A separable equation is $f(x)dx+g(y)dy=0$, so $M=f(x)$ (no $y$) and $N=g(y)$ (no $x$). Then $M_y=0=N_x$ automatically. See Separable Equations.

$F$ is unique. False ::: $F$ is unique only up to an additive constant, since $d(F+c)=dF$. This is exactly why the solution is written $F(x,y)=c$ rather than $F=0$.

$F(x,y)=c$ never cross each other. True ::: They are level curves of one function; a point cannot have two different heights, so distinct-$c$ contours are disjoint (away from singular points). See Level Curves & Contour Lines.

$M_y=N_x$ but only on a region with a hole around the origin, exactness is guaranteed. False ::: The test is only necessary there. A hole can prevent a single-valued $F$ from existing even though $M_y=N_x$ (the classic $\frac{-ydx+xdy}{x^2+y^2}$ example).

$M_y$ means differentiate $M$ with respect to $x$, since $M$ sits next to $dx$." Error ::: The subscript names the variable of differentiation, and $y$ is that variable, so $M_y=\partial M/\partial y$. The whole point of the test is to cross-differentiate: $M$ by $y$, $N$ by $x$, mirroring $F_{xy}=F_{yx}$.

$M$ in $x$, I add a constant $+C$, just like any indefinite integral." Error ::: The 'constant' of $x$-integration can still depend on $y$, so it must be $g(y)$, not a plain $C$. Dropping the $y$-dependence loses whole terms of $F$.

$F=\int Mdx+g(y)$, then sets $g^{\prime }(y)$ equal to all of $N$. Error ::: Part of $N$ already comes from ${\partial }_y\int Mdx$, which is baked into $F$. You must subtract that part first; $g^{\prime }(y)$ equals only the leftover.

$M$ in $x$ and patch anyway." Error ::: The recipe assumes a potential exists. If $M_y\ne N_x$ the patch step gives a $g^{\prime }$ that still depends on $x$ — an impossible equation for $g(y)$ — signalling failure. Find an integrating factor first.

$M=F_x$ and $N=F_y$, I can just add the two integrals: $F=\int Mdx+\int Ndy$." Error ::: That double-counts every term containing both $x$ and $y$. Those shared terms appear in both integrals, so blindly adding them gives them twice. The integrate-then-patch method exists precisely to avoid this.

$(2xy)dx+(x^2)dy=0$ I get $M_y=2x$, $N_x=2x$, so it's exact; the answer is $F_x=M\Rightarrow F=x^{2}y+3x$." Error ::: There is no $+3x$ — nothing in $M=2xy$ integrates to $3x$. Correct: $F=\int 2xydx+g(y)=x^{2}y+g(y)$, and matching $F_y=x^2=N$ gives $g^{\prime }(y)=0$, so $F=x^{2}y=c$.

mixed second partials rather than pure ones like $F_{xx}$? Because $M=F_x$ and $N=F_y$ each already carry one derivative. Differentiating $M$ by $y$ gives $F_{xy}$ and $N$ by $x$ gives $F_{yx}$ — the mixed partials are what these two paths land on, and Clairaut forces them equal.

$F(x,y)=c$ instead of solving explicitly for $y$? The differential equation only demands that we stay on a level curve of $F$; that curve is defined implicitly. Solving for $y$ is often impossible or multivalued, so the implicit form is the honest general solution.

Multiplying $Mdx+Ndy=0$ by $\mu (x,y)\ne 0$ leaves the zero set unchanged, so the same curves solve it. But $\mu M,\mu N$ can now satisfy the cross-partial test, letting the potential method work.

$dF=0$? Along a contour $F=c$ the height never changes, so its total differential $dF=F_{x}dx+F_{y}dy$ must be zero. That is exactly $Mdx+Ndy=0$ once $M=F_x,N=F_y$. See Total Differential & Partial Derivatives.

The field $(M,N)$ has potential $F$ precisely when $M=F_x,N=F_y$ — that is the definition of conservative. The test $M_y=N_x$ is the 2D curl vanishing. See Conservative Vector Fields & Potential Functions.

$0dx+0dy=0$ exact, and what is its solution? Trivially exact ($M_y=0=N_x$), with $F=$ any constant. Every point satisfies it, so there is no meaningful curve — a degenerate case where the potential is constant everywhere.

$N\equiv 0$, so the equation is $M(x,y)dx=0$, when is it exact? Exactness needs $M_y=N_x=0$, so $M$ must depend on $x$ only: $M=M(x)$. Then $F=\int Mdx$ and the "curves" are the vertical lines where $M(x)=0$ or $F(x)=c$.

linear first-order ODE $y^{\prime }+P(x)y=Q(x)$ — is it exact as written? Generally no. Written as $(P(x)y-Q(x))dx+dy=0$, we get $M_y=P(x)$, $N_x=0$, unequal unless $P\equiv 0$. Its standard integrating factor $e^{\int Pdx}$ is exactly what makes it exact.

$F_y=N$, the leftover $g^{\prime }$ still contains $x$? That is the failure signature: $g(y)$ cannot depend on $x$, so no potential exists — the equation was not exact. It confirms you must check $M_y=N_x$ before trusting the method.

Yes — $M_y=N_x$ is a pointwise condition. If it holds only on part of the plane, the potential method works only there; the full test requires it throughout a simply-connected region.