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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク integrating factor ： ウィキペディア英語版 integrating factor In mathematics, an integrating factor is a function that is chosen to facilitate the solving of a given equation involving differentials. It is commonly used to solve ordinary differential equations, but is also used within multivariable calculus when multiplying through by an integrating factor allows an inexact differential to be made into an exact differential (which can then be integrated to give a scalar field). This is especially useful in thermodynamics where temperature becomes the integrating factor that makes entropy an exact differential. == Use in solving first order linear ordinary differential equations == Integrating factors are useful for solving ordinary differential equations that can be expressed in the form :$y\text{'}+\; P(x)y\; =\; Q(x)$ The basic idea is to find some function $M(x)$, called the "integrating factor," which we can multiply through our DE in order to bring the left-hand side under a common derivative. For the canonical first-order, linear differential equation shown above, our integrating factor is chosen to be :$M(x)\; =\; e^\; P(s)\; ds\}$ In order to derive this, let $M(x)$ be the integrating factor of a first order, linear differential equation such that multiplication by $M(x)$ transforms a partial derivative into a total derivative, then: $\backslash begin$ (1) \qquad & M(x)\underset)} \\ (2) \qquad & M(x)y'+M(x)P(x)y \\ (3) \qquad & \underset} \end Going from step 2 to step 3 requires that $M(x)P(x)=M\text{'}(x)$, which is a separable differential equation, whose solution yields $M(x)$ in terms of $P(x)$: $\backslash begin$ (4) \qquad & M(x)P(x)=M'(x) \\ (5) \qquad & P(x)=\frac\\ (6) \qquad & \int P(x) dx=\ln M(x) \\ (7) \qquad & e^=M(x) \end To verify see that multiplying through by $M(x)$ gives :$y\text{'}\; e^\; P(s)\; ds\}\; +\; P(x)\; y\; e^\; P(s)\; ds\}\; =\; Q(x)e^\; P(s)\; ds\}$ By applying the product rule in reverse, we see that the left-hand side can be expressed as a single derivative in $x$ :$y\text{'}\; e^\; P(s)\; ds\}\; +\; P(x)\; y\; e^\; P(s)\; ds\}\; =\; \backslash frac(y\; e^\; P(s)\; ds\})$ We use this fact to simplify our expression to :$\backslash frac\backslash left(y\; e^\; P(s)\; ds\}\backslash right)\; =\; Q(x)\; e^\; P(s)\; ds\}$ We then integrate both sides with respect to $x$, firstly by renaming $x$ to $t$, obtaining :$y\; e^\; P(s)\; ds\}\; =\; \backslash int\_^\; Q(t)\; e^\; P(s)\; ds\}\; dt\; +\; C$ Finally, we can move the exponential to the right-hand side to find a general solution to our ODE: :$y\; =\; e^\; P(s)\; ds\}\; \backslash int\_^x\; Q(t)\; e^\; P(s)\; ds\}\; dt\; +\; Ce^\; P(s)\; ds\}$ In the case of a homogeneous differential equation, in which $Q(x)\; =\; 0$, we find that :$y\; =\; \backslash frac\; P(s)\; ds\}\}$ where $C$ is a constant. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「integrating factor」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.