Is there a way to solve an arbitrary ODE of first order? $$\frac{\text{d}y}{\text{d}x}=f(x,y)$$ I know there are some tricks to solve like separations of variables and change of variables. Especially, change of variables, how can I see what I shall substitute. But they seem to be "coincident"! Is there any algorithms to apply?
 A: Analytically, the answer is "no". Even the much simpler sub-cases $$\frac{dy}{dx}=f(x)$$ and $$\frac{dy}{dx}=f(y)$$ often don't have solutions in terms of nice analytical functions even though both are separable and the solutions to both can be expressed by using the $\int$ sign. But even for something as simple as $$\frac{dy}{dx}=e^{xy}$$ there's little chance of an analytical solution.
That said, it today's day and age, most ODEs are easily solvable numerically which is more than satisfactory for many purposes! Try http://www.wolframalpha.com/input/?i=y%27(x)+%3D+exp(x+y) and see how routinely it produces the plots of the solutions.
A: $\newcommand{\dd}{\partial}$The answer depends on what you mean by "solve". There's a general existence-uniqueness theorem for a first-order equation $y' = f(x, y)$ with $f$ Lipschitz in $y$, for example.
Your question seems to ask something more explicit, however, along the following lines. Let $P$ and $Q$ be real-valued $C^{1}$ functions in some rectangle. The first-order equation
$$
P(x, y)\, dx + Q(x, y)\, dy = 0
\tag{1}
$$
can be "integrated" or "solved" in the form $F(x, y) = c$ if it is exact, i.e., if there exists a $C^{2}$-function $F$ satisfying
$$
\frac{\dd F}{\dd x} = P,\qquad
\frac{\dd F}{\dd y} = Q.
$$
This happens if and only if an integrability conditon holds:
$$
0 = \frac{\dd Q}{\dd x} - \frac{\dd P}{\dd y}.
\tag{2a}
$$
More generally, (1) can be solved in the form $F(x, y) = c$ if there exists an integrating factor, a non-vanishing function $\mu(x, y)$ such that
$$
\frac{\dd F}{\dd x} = P\mu,\qquad
\frac{\dd F}{\dd y} = Q\mu.
$$
This happens if and only if there exists a non-vanishing function $\mu$ satisfying
\begin{align*}
0 &= \frac{1}{\mu} \left(\frac{\dd (Q\mu)}{\dd x} - \frac{\dd (P\mu)}{\dd y}\right) \\
  &= \frac{\dd Q}{\dd x} - \frac{\dd P}{\dd y} + \frac{\dd\log \mu}{\dd x} Q - \frac{\dd\log \mu}{\dd y} P.
\tag{2b}
\end{align*}
Equation (2a) is, of course, the special case $\mu = \text{const.}$
For example, the linear first-order ODE $y' + fy = g$ mentioned in your comment, written the form
$$
(fy - g)\, dx + dy = 0,\qquad
P(x, y) = f(x)y - g(x),\quad Q(x, y) = 1,
$$
is not exact, but becomes exact upon multiplication by the integrating factor
$$
\mu(x, y) = e^{\int f(x)\, dx}.
$$
