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# exact differential equation

Let $R$ be a region in $\mathbb{R}^{2}$ and let the functions $X\!:R\to\mathbb{R}$, $Y\!:R\to\mathbb{R}$ have continuous partial derivatives in $R$. The first order differential equation

$X(x,\,y)+Y(x,\,y)\frac{dy}{dx}\;=\;0$ |

or

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

is called an exact differential equation, if the condition

$\displaystyle\frac{\partial X}{\partial y}\;=\;\frac{\partial Y}{\partial x}$ | (2) |

is true in $R$.

By (2), the left hand side of (1) is the total differential of a function, there is a function $f\!:R\to\mathbb{R}$ such that the equation (1) reads

$d\,f(x,\,y)\;=\;0,$ |

whence its general integral is

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

The solution function $f$ can be calculated as the line integral

$\displaystyle f(x,\,y)\;:=\;\int_{{P_{0}}}^{P}[X(x,\,y)\,dx+Y(x,\,y)\,dy]$ | (3) |

along any curve $\gamma$ connecting an arbitrarily chosen point $P_{0}=(x_{0},\,y_{0})$ and the point $P=(x,\,y)$ in the region $R$ (the integrating factor is now $\equiv 1$).

Example. Solve the differential equation

$\frac{2x}{y^{3}}\,dx+\frac{y^{2}-3x^{2}}{y^{4}}\,dy\;=\;0.$ |

This equation is exact, since

$\frac{\partial}{\partial y}\frac{2x}{y^{3}}\;=\;-\frac{6x}{y^{4}}\;=\;\frac{% \partial}{\partial x}\frac{y^{2}-3x^{2}}{y^{4}}.$ |

If we use as the integrating way the broken line from $(0,\,1)$ to $(x,\,1)$ and from this to $(x,\,y)$, the integral (3) is simply

$\int_{0}^{x}\frac{2x}{1^{3}}\,dx+\!\int_{1}^{y}\frac{y^{2}-3x^{2}}{y^{4}}\,dy% \;=\;\frac{x^{2}}{y^{3}}-\frac{1}{y}+1\;=\;x^{2}-\frac{1}{y}+\frac{x^{2}}{y^{3% }}+1-x^{2}=\frac{x^{2}}{y^{3}}-\frac{1}{y}+1.$ |

Thus we have the general integral

$\frac{x^{2}}{y^{3}}-\frac{1}{y}\;=\;C$ |

of the given differential equation.

## Mathematics Subject Classification

34A05*no label found*

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