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