The Ambiguous Use of Differential in Solving Differential Equations? Is there any theoretical basis for employing differential methods like separation of variables in solving differential equations? As we all know, differential is formally defined as a linear transformation, so how can it be said that it's only infinitesimal and be summed up on both sides of the equation? It's not an exact explanation, let alone an exact proof.
 A: $$ f(x) = g(y)\;\frac{dy}{dx}$$
where $f$ and $g$ are known functions, and we want $y$ as a function of $x$.  So
$$ f(x)\;dx = g(y)\;dy $$
$$ F(x) = G(y) + C$$
$$ F(x) - C = G(y) $$
$$ G^{-1}(F(x) - C) = y $$
The question is how to justify this rigorously.
The answer must be the chain rule, and the question is how to show that.
If $x$ and $y$ both depend on some parameter $t$, then
$$
f(x) \frac{dx}{dt} = g(y)\frac{dy}{dt},
$$
$$
F\;'(x)\frac{dx}{dt} = G'(y)\frac{dy}{dt}.\tag{1}
$$
If $t$ happens to be $x$, this is simply the original differential equation.
The meaning of $f(x)\;dx = g(y)\;dy$ could be taken to be that no matter what variable $t$ parametrizes the graph in a smooth fashion, (1) must hold.  And the chain rule tells us that 
$$
\frac{dy/dt}{dx/dt} = \frac{dy}{dx}.
$$
Another P.O.V. might be to say that
$$
f(x)\;\Delta x \approx g(y)\;\Delta y
$$
and then examine limits as $\Delta x\to0$.
A: Assume you want to solve the Cauchy problem:
$$\tag{C} \begin{cases} y^\prime (x)= f(y(x))\ g(x) &\text{, for } x\in I \\ y(x_0)=y_0\end{cases}$$
where $I\subseteq \mathbb{R}$ an interval, $x_0\in I$ an interior point and $y_0\in \mathbb{R}$. Assume also you got existence and uniqueness for solution of (C) and some monotonicity and regularity from a priori analysis (e.g., you know any solution of (C) is strictly increasing and $C^1$). If $f(y_0)\neq 0$ then, by continuity, there is a complete neighbourhood of $x_0$ contained in $I$ s.t. $f(y(x))\neq 0$, hence you can divide both LH and RH sides in the ODE by $f(y(x))$ to get:
$$\frac{y^\prime (x)}{f(y(x))} =g(x)\; .$$
Now, let $x>x_0$. You integrate both sides of the previous equation over $[0,x]$ and, since $\tau =y(t)$ is an admissible change of variable, you gain:
$$\int_{y_0}^{y(x)} \frac{1}{f(\tau)}\ \text{d} \tau =\int_{x_0}^x\frac{y^\prime (t)}{f(y(t))}\ \text{d} t =\int_{x_0}^x g(t)\text{d} t\; ;$$
if you call $\Phi,G$ some antiderivative of $1/f,g$, then you rewrite previous equation as:
$$\Phi (y(x)) =G(x)-G(x_0)+\Phi (y_0)\; ,$$
i.e.
$$\Phi (y(x))=G(x)+c$$
with $c$ a suitable additive constant.
Notice that the latter equation is the usual way to present the solution of a separable ODE in implicit form.

For example, let us solve:
$$\begin{cases} y^\prime (x) =2x\ y(x) &\text{, in } \mathbb{R} \\ y(1)=1\; .\end{cases}$$
Picard-Lindelöf theorem can be used to prove that the problem has locally unique solution, which we can extend to a maximal interval, say $I$. Such a maximal solution $y(x)$ is of class $C^0(I)$; but then $y^\prime (x)=2x\ y(x)$ is also of class $C^0$ hence $y\in C^1(I)$; then again $y^\prime \in C^1(I)$ and $y\in C^2(I)$... Bootstrapping, we see that $y\in C^\infty (I)$. Moreover, $y$ is positive in a neighbourhood of $x_0=1$ because $y^\prime (1)=2\ y(1)=2>0$ and $y, y^\prime$ are continuous. The maximal solution $y$ cannot equal zero anywhere in $I$: in fact if by contradiction there were a point $x_1\in I$ s.t. $y(x_1)=0$, then $y$ would also solve the homogeneous Cauchy problem:
$$\begin{cases} y^\prime =x\ y \\ y(x_1)=0\end{cases}$$
which is uniquely solved by $\bar{y}(x)=0$; hence $y(x)=0$ everywhere, but this is a contradiction because $y(1)=1\neq 0$!
Therefore $y$ cannot change sign in $I$ and thus $y(x)>0$ everywhere in $I$. Consequently $y$ increases strictly in $I\cap [0,\infty[$ and decreases strictly in $I\cap ]-\infty ,0]$.
Finally, we are in the position to eveluate the solution of (C). We fix $x>1$ and compute:
$$x^2 -1 =\int_1^x 2t\ \text{d} t=\int_1^x \frac{y^\prime (t)}{y(t)}\ \text{d} t \stackrel{\tau =y (t)}{=} \int_1^{y(x)} \frac{1}{\tau}\ \text{d} \tau =\ln y(x)\; ,$$
hence:
$$\tag{1} y(x)=\exp (x^2-1)$$
for $x>1$. On the other hand, it is easily seen that $y$ as in (1) solves the ODE in (C) also for $x< 1$; therefore (1) gives the maximal solution of (C), which is defined in $I=\mathbb{R}$.
As we expected, $y$ is $C^\infty$, positive, strictly decreasing in $]-\infty ,0]$ and strictly increasing in $[0,\infty[$.
A: The explanation I used in my class today was simply that an expression
$$P(x,y) \, dx + Q(x,y) \, dy$$
has meaning only inside a line integral, and so a "differential form equation"
$$P(x,y) \, dx + Q(x,y) \, dy = 0$$
simply states that for every pair of vectors $\vec{r}_0 = (x_0, y_0)$ and $\vec{r}_1 = (x_1, y_1)$, and every smooth enough curve $C$ from one to the other, we have
$$\int_C P(x,y) \, dx + Q(x,y) \, dy = \int_C 0 = c$$
where $c$ is a constant.  
This is the key motivation.  From it, you can proceed to derive the solution rigorously as follows.  Since $C$ is not part of the data, this includes the assurance that the value of the integral must be independent of it, forcing the vector field $(P,Q)$ to be conservative and thus exact, hence the gradient $\nabla F$ of some potential function $F$ satisfying, a fortiori, the equation
$$F(x_1, y_1) - F(x_0, y_0) = c.$$
Letting $(x_1, y_1)$ vary (thus, renaming them back to $(x,y)$) and absorbing $F(x_0, y_0)$ into $c$, this gives the implicit solution
$$F(x,y) = c.$$
Now, if the field $(P,Q)$ is not exact (as seen by its not being irrotational) then you need an integrating factor to make it exact, and although this changes the differential form equation it doesn't change the associated differential equation
$$\frac{dy}{dx} = -\frac{P(x,y)}{Q(x,y)}$$
because the ratio is unchanged if $P$ and $Q$ are scaled.  This bit of logic is made rigorous using the chain rule, of course: if $F(x,y) = c$ implicitly defines $y = y(x)$ then
$$\frac{dy}{dx} = -\frac{\partial F/\partial x}{\partial F/\partial y} = -\frac{\mu P}{\mu Q}$$
if $\mu$ is the integrating factor.
