1
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ Seems odd that the variable $t$ appears in $f(t).$ Should it really be $x$ or $y$? $\endgroup$ – coffeemath Aug 15 '16 at 23:28
  • $\begingroup$ $x$ and $y$ also depend on $t$, thank you, I'm editing the question so it is clear. $\endgroup$ – Saavestro Aug 15 '16 at 23:30
  • $\begingroup$ Still looks peculiar-- if $x=x(t)$ then can $dx$ be replaced by $x'(t)dt$, and similarly from $y=y(t)$ can $dy$ be replaced by $y'(t)dt$? $\endgroup$ – coffeemath Aug 15 '16 at 23:40
  • $\begingroup$ That seems possible, still the equation is supposed to be non integrable, so it doesn't matter how you rearrenge the terms. Wat I need to know is if there is some general way of proving that in fact it is non integrable. $\endgroup$ – Saavestro Aug 16 '16 at 0:36
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.