Question

New Question !

Answer 1

Black Lives Matter

Answer 2

crap question, now looking at it properly

First thing is to check for an **exact** equation:

if \(f(x,y) = C\) then \(df = f_x ~ dx + f_y ~ dy = 0\)

But if \(f_x = x^2 - y^2\), then \(f_{xy} = - 2y\)

Whereas if \(f_y = 2xy\), then \(f_{yx} = 2y\)

And so \(f_{xy} \ne f_{yx}\)

:(

Annoying 🤔

Answer 3

So look at it as a simple ODE.

\((x^2 - y^2) ~ dx + 2 x y ~ dy = 0 \implies \dfrac{dy}{dx} = - \dfrac{x^2 - y^2}{2 x y} = - \dfrac{ \frac{x}{y} - \frac{y}{x}}{2} \)


if we invent an interesting function \(\sigma(x) = \frac{y}{x}\), then we see that \(y' = \sigma' x + \sigma x' = \sigma' x + \sigma \)

So we agree that:

\( \sigma' x + \sigma = - \dfrac{ \frac{1}{\sigma} - \sigma}{2} = - \dfrac{ 1 - \sigma^2}{2 \sigma}\)

So

\( \sigma' x = - \dfrac{ \sigma^2 + 1 }{2 \sigma}\)


And:

\( \dfrac{2 \sigma }{\sigma^2 + 1} \sigma' = - \dfrac{1 }{x}\)


Integrating both sides wrt x means that:

\( \ln |\sigma^2 + 1| = - \ln |x| + C = - (\ln |x| + \ln \alpha) = \ln \dfrac{1}{|\alpha x|}\)

So: \( \ln |\sigma^2 + 1| = \ln \dfrac{1}{|\alpha x|} \)

It sorta follows that:

\( \left(\dfrac{y}{x} \right)^2 + 1 = \dfrac{1}{\alpha x} \)

\( y^2 = x^2 \left( \dfrac{1}{\alpha x} - 1 \right) = x\left( \dfrac{1}{\alpha} - x \right) = x\left( \alpha' - x \right)\)

\(\implies y = \pm \sqrt{ x(\alpha' - x )}\)

And that's a first atempt at solving a Superman question

Answer 4

CLOSED: