# 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 $C$ may be always chosen so that any given initial condition

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

is fulfilled.

Title linear differential equation of first order LinearDifferentialEquationOfFirstOrder 2013-03-22 16:32:09 2013-03-22 16:32:09 pahio (2872) pahio (2872) 13 pahio (2872) Derivation msc 34A30 linear ordinary differential equation of first order SeparationOfVariables