47 lines
2.7 KiB
Markdown
47 lines
2.7 KiB
Markdown
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## Exact equations
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two variable equations
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$dF=\frac{ \partial F }{ \partial x }dx+\frac{ \partial F }{ \partial y }dy=0$ suppose it equals to zero (as shown in the equation) you get a horizontal plane (a constant)
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so $F(x,y)=C$
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the solution to these exact equations is given by $F()$ but how do we recover $F$ from it's partial derivatives?
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Equation of the form: $$M(x,y)dx+N(x,y)dy=0$$
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>I'm calling this #de_e_type1
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is called exact if $M(x,y)=\frac{ \partial F }{ \partial x }$ and $N(x,y)=\frac{ \partial F }{ \partial y }$ for some function $F(x,y)$
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then differentiating we get:
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$\frac{ \partial M }{ \partial y }=\frac{ \partial^{2} F }{ \partial y\partial x }$
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$\frac{ \partial N }{ \partial x }=\frac{ \partial^{2} F }{ \partial x\partial y }$ Order of going in x then y vs y then x doesn't matter as it lands you on the same point.
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We equate the two and obtain a way to check if an equation is exact:
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Exact equation$\Rightarrow \frac{ \partial M }{ \partial y }=\frac{ \partial N }{ \partial x }$ if it's continuous (?)
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also: Exact equation$\Leftarrow \frac{ \partial M }{ \partial y }=\frac{ \partial N }{ \partial x }$
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Test for exactness:
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exact $\iff \frac{ \partial M }{ \partial y }=\frac{ \partial N }{ \partial x }$ (this can be proved, but it wasn't proved in class)
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#end of lecture 4
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#start of lecture 5
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last lecture we talked about exact equations
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We only knew about N and M which are the partials of F()
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so how do we recover F?
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between N and M, choose the one that is easier to integrate. Let's choose M.
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$M=\frac{ \partial F }{ \partial x }$
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$F(x,y)=\int M(x,y) \, dx$
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$F(x,y)=\int M(x,y) \, dx+g(y)$ 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.
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now 2nd condition: $N=\frac{ \partial F }{ \partial y }=\frac{ \partial }{ \partial y }\int M(x,y) \, dx+g'(y)=N(x,y)$
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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.
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#ex #de_e_type1
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$$\underbrace{( 2xy+3 )}_{ M }dx+\underbrace{ (x^2-1) }_{N}dy=0$$
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$\frac{ \partial M }{ \partial y }=2x=\frac{ \partial N }{ \partial x }=2x$ so its exact!
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$\frac{ \partial F }{ \partial y }=N(x,y)=x^2-1$
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integrate $N(x,y)$ wrt to y:
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$F(x,y)=(x^2-1)y+g(x)$ (side note: although we say g is any function, it should be differentiable tho)
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$\frac{ \partial F }{ \partial x }=M(x,y)=2xy+3=2xy+g'(x)$
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$g(x)=3x+C_{1}$
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$F(x,y)=(x^2-1)y+g(x)\Rightarrow F(x,y)=(x^2-1)y+3x=C_{2}-C_{1}=C$
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We are done:
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$$(x^2-1)y+3x=C$$
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there is also another method to solve exact equations (see Wikipedia article, but the prof says this method is easier, I believe him) |