Integrating factor

Description

An Exact differential equation is easy to solve but they are very rare. However, sometimes it is possible to modify them by an integrating factor, making them exact.

Definition

Let $G\subseteq {\mathbb{R}}^2$ be a Gebiet and $f,g:G\to \mathbb{R}$ Partially differentiable and continous. For an Exact differential equation f(t, x) + g(t, x)\dot x = 0 \tag{1} if there exists a Stetig Differenzierbare Funktion $m:G\to \mathbb{R}\backslash \{0\}$, making the differential equation m(t, x)f(t, x) + m(t, x)g(t, x)\dot x = 0 \tag{2} exact, then $m$ is called an integrating factor (or Euler multiplicator) of the ODE.

Since $m$ is never zero $\lambda$ solves $(1)$ if and only if it solves $(2)$.

Properties

Tip

Examples

Example

Application

Sometimes it can be very hard to find an integral factor. We present one way which works most of the time. Note that if $m(t,x)f(t,x)+m(t,x)g(t,x)\dot{x}=0$ should be exact, then it also need to fulfill ${\partial }_x(m(t,x)f(t,x))={\partial }_t(m(t,x)g(t,x))$. Writing $m$ as the concatenation of two functions $M(u)$ yielding $\frac{M^{\prime }(u)}{M(u)}=\frac{{\partial }_{x}f(t,x)-{\partial }_{t}g(t,x)}{g(t,x){\partial }_{t}u-f(t,x){\partial }_{x}u}$ Which is hopefully an differential equation that can be solved.

Calculating an integral factor

Let $f(t,x)+g(t,x)\dot{x}=0$ be an inexact differential equation. To find an integrating factor $m(t,x)$ we do the following:

1. Choose some function $u(t,x)$ (like $t$, $x$, $t\pm x$, $tx$) and calculate $\frac{M^{\prime }}{M}=\frac{{\partial }_{x}f-{\partial }_{t}g}{g{\partial }_{t}u-f{\partial }_{x}u}$ The goal is that all variables $t,x$ except $u$ disappear.
2. If $\frac{M^{\prime }(t,x)}{M(t,x)}$ is a function in $u$ then solve by $M(u)=e^{\int H(u)du}$. The integrating factor is then given by $m(x,t)=M(u(t,x))$.

I feel like, this is way to complicated. It might be easier to just guess the factor. In most cases it will be something like $e^{2t}$ or $e^{-t}$.