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Homemethod of integrating factors

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# method of integrating factors

The method of integrating factors is in principle a means for solving ordinary differential equations of first order. It has not great practical significance, but is theoretically important.

Let us consider a differential equation solved for the derivative $y^{{\prime}}$ of the unknown function and write the equation in the form

$\displaystyle X(x,\,y)\,dx+Y(x,\,y)\,dy\;=\;0.$ | (1) |

We assume that the functions $X$ and $Y$ have continuous partial derivatives in a region $R$ of $\mathbb{R}^{2}$.

If there is a solution of (1) which may be expressed in the form

$f(x,\,y)\;=\;C$ |

with $f$ having continuous partial derivatives in $R$ and with $C$ an arbitrary constant, then it’s not difficult to see that such an $f$ satisfies the linear partial differential equation

$\displaystyle X\frac{\partial f}{\partial y}-Y\frac{\partial f}{\partial x}\;=% \;0.$ | (2) |

Conversely, every non-constant solution $f$ of (2) gives also a solution $f(x,\,y)=C$ of (1). Thus, solving (1) and solving (2) are equivalent tasks.

It’s straightforward to show that if $f_{0}(x,\,y)$ is a non-constant solution of the equation (2), then all solutions of this equation are $F(f_{0}(x,\,y))$ where $F$ is a freely chosen function with (mostly) continuous derivative.

The connection of the equations (1) and (2) may be presented also in another form. Suppose that $f(x,\,y)=C$ is any solution of (1). Then (2) implies the proportion equation

$\frac{f_{x}^{{\prime}}}{X}\;=\;\frac{f_{y}^{{\prime}}}{Y}.$ |

If we denote the common value of these two ratios by $\mu(x,\,y)=\mu$, then we have

$f_{x}^{{\prime}}\;=\;\mu X,\qquad f_{y}^{{\prime}}\;=\;\mu Y.$ |

This gives to the differential of the function $f$ the expression

$d\,f(x,\,y)\;=\;\mu(x,\,y)(X(x,\,y)\,dx+Y(x,\,y)\,dy).$ |

We see that $\mu(x,\,y)$ is the integrating factor or Euler multiplicator of the given differential equation (1), i.e. the left hand side of (1) turns, when multiplied by $\mu(x,\,y)$, to an exact differential.

Conversely, any integrating factor $\mu$ of (1), i.e. such that $\mu X\,dx+\mu Y\,dy$ is the differential of some function $f$, is easily seen to determine the solutions of the form $f(x,\,y)=C$ of (1). Altogether, solving the differential equation (1) is equivalent with finding an integrating factor of the equation.

When an integrating factor $\mu$ of (1) is available, the solution function $f$ can be gotten from the line integral

$f(x,\,y)\;=:\;\int_{{P_{0}}}^{P}[\mu(x,\,y)X(x,\,y)\,dx+\mu(x,\,y)Y(x,\,y)\,dy]$ |

along any curve $\gamma$ connecting an arbitrarily chosen point $P_{0}=(x_{0},\,y_{0})$ and the point $P=(x,\,y)$ in the region $R$.

Note. In general, it’s very hard to find a suitable integrating factor. One special case where such can be found, is that $X$ and $Y$ are homogeneous functions of same degree: then the expression $\displaystyle\frac{1}{xX+yY}$ is an integrating factor.

Example. In the differential equation

$(x^{4}+y^{4})\,dx-xy^{3}\,dy\;=\;0$ |

we see that $X=:x^{4}+y^{4}$ and $Y=:-xy^{3}$ both define a homogeneous function of degree 4. Thus we have the integrating factor $\displaystyle\mu=:\frac{1}{x^{5}+xy^{4}-xy^{4}}=\frac{1}{x^{5}}$, and the left hand side of the equation

$\left(\frac{1}{x}+\frac{y^{4}}{x^{5}}\right)\,dx-\frac{y^{3}}{x^{4}}\,dy\;=\;0$ |

is an exact differential. We can integrate it along the broken line, first from $(1,\,0)$ to $(x,\,0)$ and then still to $(x,\,y)$, obtaining

$f(x,\,y)\;=:\;\int_{1}^{x}\left(\frac{1}{x}+\frac{0^{4}}{x^{5}}\right)\,dx-% \int_{0}^{y}\frac{y^{3}\,dy}{x^{4}}\;=\;\ln|x|-\frac{y^{4}}{4x^{4}}.$ |

So the general solution of the given differential equation is

$\ln|x|-\frac{y^{4}}{4x^{4}}\;=\;C.$ |

# References

- 1 E. Lindelöf: Differentiali- ja integralilasku III 1. Mercatorin Kirjapaino Osakeyhtiö, Helsinki (1935).

## Mathematics Subject Classification

35-00*no label found*34-00

*no label found*

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