Differential equations of the form $F(x,y')=0$ I have issues with understanding with such functions
For example, I was given to solve this differential equation:
$$
y'=e^{y'}-x
$$
In this equation I have to "taking into account the form of the function F, one of these functions is specified (chosen), and the other is found taking into account the identity".
I'm not really sure what it means and how to use this in order to solve the equation
If someone can explain me and give an approach in solving such equation I'll be really appreciated!
 A: $$y^{''}x = -y^{''}y^{'} + y^{''}e^{y^{'}}$$
Integrating both sides
$$ \int y^{''}x \,dx =  - \frac{1}{2}y^{'} + e^{y^{'}}$$
Integration by parts
$$ y^{'}x - y =  - \frac{1}{2}y^{'} + e^{y^{'}} + A$$
Now with 2 equations we can eliminate $e^{y^{'}}$
$$ y^{'}x - y =  - \frac{1}{2}y^{'} + (y^{'} + x) + A$$
$$  - y =  y^{'}( \frac{1}{2} - x) + x + A$$
$$  0 =  y^{'} + \frac{y}{( \frac{1}{2} - x)} + \frac{x+A}{( \frac{1}{2} - x)} $$
$$  \frac{d}{dx}\left(\frac{y}{ \frac{1}{2} - x} + B\right) = -\frac{x+A}{( \frac{1}{2} - x)^2}$$
The rhs hand side can be easily integrated and then solved for y.
You can then plug back into original equation and check over which region of the parameters B and C the ODE is satisfied.
The reason the method works is because you need to eliminate $e^{y^{'}$ to have any hope of solving the ODE.
A: There is no solution for the general case. The method of integration by differentiation can be of use in some cases. Defining $y'=\xi$ we arrive at an equation solved for the dependent variable's derivative,
\begin{align}
F(x,\xi)=0,\quad \text{and}\quad \frac{\mathrm dy}{\mathrm d\xi}=\frac{-\xi F_\xi}{F_x}.
\end{align}
The solution to this equation would yield a parametric solution. In the case that the equation can solved for $x$, i.e.
\begin{align}
x=f(y'),
\end{align}
our method yields the solution
\begin{align}
x=f(\xi),\quad y=\int\xi f'(\xi)\mathrm d\xi +C.
\end{align}
For your example we find the solution
\begin{align}
x=\xi-e^\xi,\quad y=\frac{\xi^2}{2}+(1-\xi)e^\xi+C.
\end{align}
A: For your specific problem
$$y'=e^{y'}-x$$   as @MachineLearner showed, you can solve for $y'$ and obtain
$$y'=-W\left(-e^{-x}\right)-x$$ where $W(.)$ is Lambert function.
Integrating, you should obtain
$$y=\frac{1}{2}
   W\left(-e^{-x}\right)^2+W\left(-e^{-x}\right)-\frac{1}{2}x^2+C$$
A: To help understanding the function, I recommend looking at
$$ f'(x)=\lim_{\Delta x\rightarrow 0}{\frac{f(x+\Delta x)-f(x)}{\Delta  x}}$$, but with a small tweak where $\Delta x=1$, and  then your differential equation becomes the following:
$$ y(x+1)-y(x) = e^{y(x+1)-y(x)}-x$$, i.e. when you have a sequence of numbers y(0),y(1),y(2),y(3), and x=0,x=1,x=2,x=3, you can get:
$$ y(1)-y(0) = e^{y(1)-y(0)}-0$$, and
$$ y(2)-y(1) = e^{y(2)-y(1)}-1$$ etc...
Each of these equations you have two unknowns, for example y(1) and y(0), but just one equation, so if you do y(0)=C, then it magically can be solved as normal equation solving.
And from that, you can get the whole sequence y(0),y(1),y(2),y(3)... (except for the C)
Then moving from $\Delta x=1$ to $\Delta x \rightarrow 0$ world is left as excersize for the reader.
2nd excersize for the reader is to figure out how to get polynomial which goes through all points (0,y(0)),(1,y(1)),(2,y(2)),...
