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Exact equations

two variable equations

$dF=\frac{ \partial F }{ \partial x }dx+\frac{ \partial F }{ \partial y }dy=0$ 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_exact

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_exact

$$\underbrace{( 2xy+3 )}_{ M }dx+\underbrace{ (x^2-1) }_{N}dy=0$$

$\frac{ \partial M }{ \partial y }=2x=\frac{ \partial N }{ \partial x }=2x$ 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)

$\frac{ \partial F }{ \partial x }=M(x,y)=2xy+3=2xy+g'(x)$

$g(x)=3x+C_{1}$

$F(x,y)=(x^2-1)y+g(x)\Rightarrow F(x,y)=(x^2-1)y+3x=C_{2}-C_{1}=C$

We are done:

$$(x^2-1)y+3x=C$$

there is also another method to solve exact equations (see Wikipedia article, but the prof says this method is easier, I believe him)

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