I have the following differential equation:
$y(t)dx+(f(t)-x(t))dy=0$
It is suppose to be non integrable for a differentiable but arbitrary $f(t)$. How do I know this is true?
This is part of the problem 1.6 in Goldstein's Classical Mechanics 3rd edition book. Also, in the text it says that, in principle, an integrating factor always can be found for any first order differential equation that involves only two variables. How can I prove this statement?
This is a comment, but too long to be put in the comments section.
$$y(t)dx+\left(f(t)-x(t)\right)dy=0$$ $$\begin{cases} f(t) \text{ is a given (known) function.}\\ x(t) \text{ and } y(t) \text{ are unknown functions.} \end{cases}$$ It is expected to express the unknown functions $x$ and $y$ in terms of the known function $f$.
This cannot be done because there is only one equation , but two unknowns : The problem is undetermined.
This means that we can find as many solutions as we want, for example in defining $x(t)$ as any given differentiable function. In other words, if $f(t)$ and $x(t)$ are two different given functions, then the initial equation is integrable and the solution is: $$y(t)=e^{\int \frac{x'(t)}{x(t)-f(t)}dt}$$ Since $x(t)$ is given, $x'(t)$ is known. Thus the integral is well defined.