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Homelinear differential equation of first order

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# linear differential equation of first order

An ordinary linear differential equation of first order has the form

$\displaystyle\frac{dy}{dx}+P(x)y\;=\;Q(x),$ | (1) |

where $y$ means the unknown function, $P$ and $Q$ are two known continuous functions.

For finding the solution of (1), we may seek a function $y$ which is product of two functions:

$\displaystyle y(x)\;=\;u(x)v(x)$ | (2) |

One of these two can be chosen freely; the other is determined according to (1).

We substitute (2) and the derivative $\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$ in (1), getting $u\frac{dv}{dx}+v\frac{du}{dx}+Puv=Q$, or

$\displaystyle u\left(\frac{dv}{dx}+Pv\right)+v\frac{du}{dx}\;=\;Q.$ | (3) |

If we chose the function $v$ such that

$\frac{dv}{dx}+Pv\;=\;0,$ |

this condition may be written

$\frac{dv}{v}\;=\;-P\,dx.$ |

Integrating here both sides gives $\ln{v}=-\int P\,dx$ or

$v\;=\;e^{{-\int Pdx}},$ |

where the exponent means an arbitrary antiderivative of $-P$. Naturally, $v(x)\neq 0$.

Considering the chosen property of $v$ in (3), this equation can be written

$v\frac{du}{dx}\;=\;Q,$ |

i.e.

$\frac{du}{dx}\;=\;\frac{Q(x)}{v(x)},$ |

whence

$u\;=\;\int\frac{Q(x)}{v(x)}\,dx+C\;=\;C+\!\int Qe^{{\int Pdx}}dx.$ |

So we have obtained the solution

$\displaystyle y\;=\;e^{{-\int P(x)dx}}\left[C+\!\int Q(x)e^{{\int P(x)dx}}dx\right]$ | (4) |

of the given differential equation (1).

The result (4) presents the general solution of (1), since the arbitrary constant $C$ may be always chosen so that any given initial condition

$y\;=\;y_{0}\quad\mathrm{when}\quad x\;=\;x_{0}$ |

is fulfilled.

## Mathematics Subject Classification

34A30*no label found*

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