Question bank — Integrating factors for non-exact equations

True in principle — an integrating factor always exists locally for a smooth $Mdx+Ndy$, but it may depend on both $x$ and $y$ and be as hard to find as solving the ODE directly.

True — the constant function $1$ is a perfectly good $\mu$; an exact equation is just the special case where no rescaling is needed.

False — $\mu$ must be nonzero; multiplying by $0$ turns everything into $0=0$ and destroys the solution curves entirely.

False — since $\mu \ne 0$, the equation $\mu Mdx+\mu Ndy=0$ vanishes on exactly the same set as before, so the solution curves are identical.

True — this happens (e.g. it can also admit $\mu (xy)$), and both make it exact; the resulting potentials differ but describe the same solution family.

True — that ratio equals ${\mu }^{\prime }/\mu$, a pure function of $x$, which integrates to a genuine $\mu (x)=\exp \int gdx$.

False — it still has solutions; you just cannot use the two easy single-variable formulas and must try other ansätze like $\mu (xy)$ or $\mu (x+y)$.

False — it is exact only if the ratio truly was single-variable; if you forced the formula on an invalid ratio, $M_y^{\ast }\ne N_x^{\ast }$ still.

The error is stopping after the $x$-test; you must also try the $y$-test $\frac{N_x-M_y}{M}=\frac{1}{y}$, which is pure $y$ and gives $\mu =y$.

Wrong numerator sign — the $y$-case is $\frac{N_x-M_y}{M}$; the minus comes from $-M{\mu }^{\prime }=\mu (M_y-N_x)$ moving across.

You must multiply the whole equation, both $M$ and $N$, by $\mu$; scaling one term breaks the equation's meaning and won't be exact.

The integration 'constant' is $\varphi (y)$, not a true constant $C$; you must differentiate in $y$ and match $N^{\ast }$ to pin it down before writing $F=C$.

$\mu =\exp (\int \frac{1}{x}dx)=\exp (\ln x)=x$, not $\frac{1}{x}$; forgetting the exponential (and integrating instead of exponentiating) is the classic slip.

The formula guarantees exactness only if the ratio was genuinely $x$-only; a re-check $M_y^{\ast }=N_x^{\ast }$ catches arithmetic errors that quietly poison the ratio.

Because ${\mu }_y=0$ kills the $M{\mu }_y$ term, leaving $N{\mu }^{\prime }=\mu (M_y-N_x)$, a separable ODE in $x$ alone rather than a two-variable PDE.

The left side ${\mu }^{\prime }/\mu$ is a function of $x$; it can only equal the right side if the right side also has no $y$, otherwise the equation is self-contradictory.

Exactness means $Mdx+Ndy=dF$, so the ODE reads $dF=0$; a quantity with zero total differential is constant along solutions, hence $F=C$.

Writing $y^{\prime }+Py=Q$ as $(Py-Q)dx+dy=0$ has $\frac{M_y-N_x}{N}=P(x)$, pure $x$, so the general formula reproduces exactly the linear factor.

Because $M=F_x,N=F_y$ forces $M_y=F_{xy}$ and $N_x=F_{yx}$, which are equal only when mixed partials commute for smooth $F$.

$Mdx+Ndy=dF$ means $(M,N)$ is the gradient of $F$, so it is a conservative field with potential $F$ and zero curl $M_y-N_x=0$.

Because the general $\mu (x,y)$ satisfies a PDE as hard as the original problem; restricting to one variable makes it a solvable ODE, which is the whole practical payoff.

No — $M_y=N_x$ means it is already exact, so $\mu =1$; go straight to finding $F$ regardless of surface complexity.

No — $\frac{M_y-N_x}{N}$ divides by $N$, so $N\equiv 0$ makes it undefined; but $N=0$ means $Mdx=0$, which is trivially separable instead.

Fine — a constant is a valid single-variable function, giving $\mu =e^{cx}$ or $e^{cy}$; constants are the easiest allowed case, not a failure.

It means $M_y=N_x$, so $g(x)=0$ and $\mu =\exp (0)=1$; the formula gracefully returns "already exact," confirming consistency.

Yes — at $x=0$ the factor vanishes, so exactness (and the potential's validity) can break on that line; solutions crossing $x=0$ may need separate handling since $\mu \ne 0$ was assumed.

Yes — methods are not exclusive; if an equation is separable it may also be exact after some $\mu$, and either route reaches the same solution family.

The formula $\frac{N_x-M_y}{M}$ divides by $M=0$ and is undefined; but $M=0$ gives $Ndy=0$, again a degenerate/separable case handled directly.