AKA 완전미분방정식 완미방 exact_equation
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완전미분exact_differential
전미분,total_derivative =
전미분,total_derivative
전미분,total_differential =전미분,total_differential
완전미분exact_differential
전미분,total_derivative =
전미분,total_derivative전미분,total_differential =
전미분,total_differential$\displaystyle f(x,y)=c$ 가 있을 때
그것의 total_differential $\displaystyle \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy=0$
의 solution 이 ...see 2021-03-16 15m again and tbw
그것의 total_differential $\displaystyle \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy=0$
의 solution 이 ...see 2021-03-16 15m again and tbw
(KU안춘기)
Exact Equations
Given
상수,constant)
This means the solution for $\displaystyle M(x,y)dx+N(x,y)dy=0$ is $\displaystyle f(x,y)=c$
(Beelee 2022-03-17)
$\displaystyle M(x,y)dx+N(x,y)dy=0$
and it fulfills$\displaystyle {\partial M \over \partial y}={\partial N \over \partial x}$
then it is an 완전미분exact_differential of some function $\displaystyle f(x,y)=c$ (c는 상수,constant)This means the solution for $\displaystyle M(x,y)dx+N(x,y)dy=0$ is $\displaystyle f(x,y)=c$
(Beelee 2022-03-17)
다음(differential form) 방정식
(Schaum DE 1.11. 다만 Ch2에서 more precise definition of exactness is given이라는 걸로 보아 완벽한 정의는 아님)
$\displaystyle M(x,y)dx+N(x,y)dy=0$
이 다음 조건을 만족하면$\displaystyle \frac{\partial M(x,y)}{\partial y}=\frac{\partial N(x,y)}{\partial x}$
exact라고 한다.(Schaum DE 1.11. 다만 Ch2에서 more precise definition of exactness is given이라는 걸로 보아 완벽한 정의는 아님)
다음 꼴의 일계미방
(Zill 6e Definition 2.4.1에 의하면)
$\displaystyle M(x,y)dx+N(x,y)dy=0$
is said to be exact equation if the expression on the left side is an exact differential(완전미분exact_differential).(Zill 6e Definition 2.4.1에 의하면)
$\displaystyle M(x,y)+N(x,y)y'=0$
$\displaystyle M(x,y)dx+N(x,y)dy=0$ (differential form)
$\displaystyle M(x,y)dx+N(x,y)dy=0$ (differential form)
(O'Neil 7e 1.3)
$\displaystyle Q(x,y)\frac{dy}{dx}+P(x,y)=0$ is an exact equation iff
the differential form $\displaystyle Q(x,y)dy+P(x,y)dx$ is exact,
i.e. there exists a function $\displaystyle f(x,y)$ for which
the differential form $\displaystyle Q(x,y)dy+P(x,y)dx$ is exact,
i.e. there exists a function $\displaystyle f(x,y)$ for which
$\displaystyle df=Q(x,y)dy+P(x,y)dx$
(Cambridge class notes에서)ex ¶
ex. (from Zill ex1)
$\displaystyle 2xydx+(x^2-1)dy=0$
sol.
With $\displaystyle M(x,y)=2xy,N(x,y)=x^2-1$ we have
따라서 양함수꼴은 x가 1이나 -1이 아닐 때 $\displaystyle y=c/(x^2-1)$ 이다.
$\displaystyle 2xydx+(x^2-1)dy=0$
sol.
With $\displaystyle M(x,y)=2xy,N(x,y)=x^2-1$ we have
$\displaystyle \frac{\partial M}{\partial y}=2x=\frac{\partial N}{\partial x}$
그래서 exact하다. 그리하여... $\displaystyle \exists f(x,y)$ such that$\displaystyle \frac{\partial f}{\partial x}=2xy\textrm{ and }\frac{\partial f}{\partial y}=x^2-1.$
From the first of these equations we obtain, after integrating, (???)$\displaystyle f(x,y)=x^2y+g(y)$
여기서 y에 대해 편미분을 하고 and setting the result equal to N(x,y) gives$\displaystyle \frac{\partial f}{\partial y}=x^2+g'(y)=x^2-1.\;\leftarrow N(x,y)$
It follows that$\displaystyle g'(y)=-1$ and
$\displaystyle g(y)=-y$
따라서, $\displaystyle f(x,y)=x^2y-y$ 이고 방정식의 해의 음함수꼴은 $\displaystyle x^2y-y=c$ 이다.$\displaystyle g(y)=-y$
따라서 양함수꼴은 x가 1이나 -1이 아닐 때 $\displaystyle y=c/(x^2-1)$ 이다.
해는 $\displaystyle f(x,y)=x^2y-y$ 가 아니다.
2021-04-04
완전미방 풀이법.
먼저 $\displaystyle \frac{\partial f}{\partial x}=M(x,y)$ 양변을 $\displaystyle x$ 에 관해 적분하면
해 $\displaystyle f(x,y)=c$ 를 얻는다.
완전미방 풀이법.
$\displaystyle M(x,y)dx+N(x,y)dy=0$
가 완전미방이면,$\displaystyle \frac{\partial f}{\partial x}=M(x,y),\;\frac{\partial f}{\partial y}=N(x,y)$
를 만족하는 $\displaystyle f(x,y)$ 가 존재한다.먼저 $\displaystyle \frac{\partial f}{\partial x}=M(x,y)$ 양변을 $\displaystyle x$ 에 관해 적분하면
$\displaystyle f(x,y)=\int M(x,y)dx+g(y)$ ......(1)
(1)을 $\displaystyle y$ 에 관해 편미분하면$\displaystyle \frac{\partial f}{\partial y}=\frac{\partial}{\partial y}\int M(x,y)dx+g'(y)$ 에서
$\displaystyle g'(y)=N(x,y)-\frac{\partial}{\partial y}\int M(x,y)dx$ ......(2)
(2)를 $\displaystyle y$ 에 관해 적분하여 (1)에 대입하면$\displaystyle g'(y)=N(x,y)-\frac{\partial}{\partial y}\int M(x,y)dx$ ......(2)
해 $\displaystyle f(x,y)=c$ 를 얻는다.
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