$dF=\frac{ \partial F }{ \partial x }dx+\frac{ \partial F }{ \partial y }dy=0$ suppose it equals to zero (as shown in the equation) you get a horizontal plane (a constant)
so $F(x,y)=C$
the solution to these exact equations is given by $F()$ but how do we recover $F$ from it's partial derivatives?
Equation of the form: $$M(x,y)dx+N(x,y)dy=0$$
>I'm calling this #de_e_type1
is called exact if $M(x,y)=\frac{ \partial F }{ \partial x }$ and $N(x,y)=\frac{ \partial F }{ \partial y }$ for some function $F(x,y)$
then differentiating we get:
$\frac{ \partial M }{ \partial y }=\frac{ \partial^{2} F }{ \partial y\partial x }$
$\frac{ \partial N }{ \partial x }=\frac{ \partial^{2} F }{ \partial x\partial y }$ Order of going in x then y vs y then x doesn't matter as it lands you on the same point.
We equate the two and obtain a way to check if an equation is exact:
Exact equation$\Rightarrow \frac{ \partial M }{ \partial y }=\frac{ \partial N }{ \partial x }$ if it's continuous (?)
also: Exact equation$\Leftarrow \frac{ \partial M }{ \partial y }=\frac{ \partial N }{ \partial x }$
Test for exactness:
exact $\iff \frac{ \partial M }{ \partial y }=\frac{ \partial N }{ \partial x }$ (this can be proved, but it wasn't proved in class)
#end of lecture 4
#start of lecture 5
last lecture we talked about exact equations
We only knew about N and M which are the partials of F()
so how do we recover F?
between N and M, choose the one that is easier to integrate. Let's choose M.
$M=\frac{ \partial F }{ \partial x }$
$F(x,y)=\int M(x,y) \, dx$
$F(x,y)=\int M(x,y) \, dx+g(y)$ where g is any function of y. The constant of integration may depend on y because if you undo by differentiating with respect to x the term would still disappear.
now 2nd condition: $N=\frac{ \partial F }{ \partial y }=\frac{ \partial }{ \partial y }\int M(x,y) \, dx+g'(y)=N(x,y)$
to reiterate, first test if equation is exact, then take m or n and integrate with x or y respectively then differentiate with respect to y or x respectively.
#ex #de_e_type1
$$\underbrace{( 2xy+3 )}_{ M }dx+\underbrace{ (x^2-1) }_{N}dy=0$$
$\frac{ \partial M }{ \partial y }=2x=\frac{ \partial N }{ \partial x }=2x$ so its exact!
$\frac{ \partial F }{ \partial y }=N(x,y)=x^2-1$
integrate $N(x,y)$ wrt to y:
$F(x,y)=(x^2-1)y+g(x)$ (side note: although we say g is any function, it should be differentiable tho)
$\frac{ \partial F }{ \partial x }=M(x,y)=2xy+3=2xy+g'(x)$
