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$$\frac{\text{d}y}{\text{d}x}=f(x,y)$$ 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?

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    $\begingroup$ There are no analytic methods that always work. $\endgroup$
    – almagest
    Apr 18, 2016 at 13:20

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$\newcommand{\dd}{\partial}$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 $$ \mu(x, y) = e^{\int f(x)\, dx}. $$

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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.

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  • $\begingroup$ OK. E.g. a linear- first order ODE: $y'+fy=g$, can be 'solved' arbitrarily: $y=\frac{\int{e^{\int f dt}g dt}+C}{e^{\int f dt}}$. I didn't expect an closed form. E.g. $y'=e^{-x^2}$ Of course, we don't have elementary functions as solution, but I can easily write integral. So you meant, also, even if I search such kind of formula, there won't be any, either? $\endgroup$
    – Upc
    Apr 18, 2016 at 17:15

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