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:

$y(x)=u(x)v(x)$ (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

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

If we chose the function $v$ such that

$$\frac{dv}{dx}+Pv=\mathrm{\hspace{0.33em}0},$$

this condition may be written

$$\frac{dv}{v}=-Pdx.$$

Integrating here both sides gives  $\ln v=-\int P𝑑x$  or

$$v=e^{-\int P𝑑x},$$

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

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

$$u=\int \frac{Q(x)}{v(x)}𝑑x+C=C+\int Q{e}^{\int P𝑑x}𝑑x.$$

So we have obtained the solution

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

of the given differential equation (1).

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

$$y=y_0\mathit{\hspace{1em}}\mathrm{when}\mathit{\hspace{1em}}x=x_0$$

is fulfilled.

Title	linear differential equation of first order
Canonical name	LinearDifferentialEquationOfFirstOrder
Date of creation	2013-03-22 16:32:09
Last modified on	2013-03-22 16:32:09
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	13
Author	pahio (2872)
Entry type	Derivation
Classification	msc 34A30
Synonym	linear ordinary differential equation of first order
Related topic	SeparationOfVariables