Question

Integrating Factors for Exact Differential equations;
Why do they work,

Answer 1

@UnkleRhaukus where t'is da questyon

Answer 2

i can answer solve them but i dont understand why they method works

Answer 3

\[\frac{dy}{dx}=\frac x{x^2y+y^3}\]

Answer 4

it might have to do with the nature of the function that they came from ... im thinking (prolly wrong tho) that convergence of dx and dy plays a part; or some intricate play between them.

Do you recall how to find a saddle point?

Answer 5

something partials negative ~, am i close

Answer 6

yeah, i believe its like
\[F_{xx}F_{yy}-F^2_{xy}\]

Answer 7

if Fxy = Fyx we have an exact diffyQ right?

Answer 8

right,

Answer 9

i have a thought on the tip of my brain, but its just not forming into something cohesive yet :)

Answer 10

you've shed some lights in the right direction,

Answer 11

with any luck, someone smarter than me can fill in some gaps ... or point it in a righter direcetion :) ill have to review this for prolly abt the rest of the week now :/

Answer 12

\[\newcommand
\p \newcommand
\p \dd [1] { \,\mathrm d#1 } % infinitesimal
\p \de [2] { \frac{ \mathrm d #1}{\mathrm d#2} } % first order derivative
\p \pa [2] { \frac{ \partial #1}{\partial #2} } % first order partial
\begin{align*} \de yx &=\frac x{x^2y+y^3} \\
(-x)\dd x +(x^2y+y^3)\dd y &=0 \\
\\
\pa My=0\qquad\pa Nx=2xy&\qquad\text{ not exact} \\
\\
\frac{N_x-M_y}M=\frac{2xy}{-x}=-2y \\
&&R(y)
&=\exp\int-2y\dd y \\
&& &=e^{-y^2} \\
(-xe^{-y^2})\dd x +(x^2ye^{-y^2}+y^3e^{-y^2})\dd y &=0 \\
\\
\pa {\overline M}y=2xye^{-y^2}\qquad\pa {\overline N}x=&2xe^{-y^2}\qquad\text{ exact} \\
% \\
% f &=\int-xe^{-y^2}\dd x \\
% &=-\tfrac12x^2e^{-y^2}+g(y) \\
% \\
% \pa fy&= \\
\end{align*}\]

Answer 13

\[M_{yy}=0\qquad\qquad N_{xx}=2y\]

Answer 14

*******\[\frac{\partial \overline {M }}y=2xye^{-y^2}\qquad\frac{\partial \overline N}{\partial x}=2xye^{-y^2}\qquad\text{ exact} \\\]

Answer 15

\[\overline M_{yy}=2xe^{-y^2}-4xy^2e^{-y^2}\qquad\qquad \overline N_{xx}=2ye^{-y^2}\] ?

Answer 16

\[\overline M_{xy}=2xe^{-y^2}\qquad\qquad \overline N_{yx}=2xe^{-y^2}-4xy^2e^{-y^2}\]

Answer 17

\[\overline N_{xx}\overline M_{yy}−\overline M_{xy}\overline N_{yx}=0\]

Answer 18

...

So inconclution integrating factors work by turning a not exact equation into a exact equation by somehow something saddle points

Answer 19

lol