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.