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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'+ 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:
\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'(x), which is a separable differential equation, whose solution yields M(x) in terms of P(x):
\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' 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' e^ P(s) ds} + P(x) y e^ P(s) ds} = \frac(y e^ P(s) ds})
We use this fact to simplify our expression to
:\frac\left(y e^ P(s) ds}\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} = \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} \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 = \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)
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