# Calculus Course/Differential Equations/1st Order Differential Equations

## 1st Order Differential Equations

A 1st order differential equation has the form shown below

${\displaystyle A{\frac {d}{dx}}f(x)+B=0}$

It can be shown that roots o the differential equation above is

${\displaystyle f(x)=Ae^{(}-\alpha x)}$
${\displaystyle \alpha ={\frac {B}{A}}}$

## Proof

The above equation can be rewritten as

${\displaystyle {\frac {df(x)}{dx}}+{\frac {B}{A}}f(x)=0}$

Then

${\displaystyle {\frac {df(x)}{dx}}=-{\frac {B}{A}}f(x)}$
${\displaystyle \int {\frac {df(x)}{f(x)}}=-{\frac {B}{A}}\int dx}$
${\displaystyle Lnf(x)=-{\frac {B}{A}}x+C}$
${\displaystyle f(x)=e^{(}-{\frac {B}{A}}x+C)}$
${\displaystyle f(x)=Ae^{(}-{\frac {B}{A}}x)}$

## Summary

First ordered differential equation of the form

${\displaystyle A{\frac {d}{dx}}f(x)+B=0}$

has a exponential root of the form

${\displaystyle f(x)=e^{(}-\alpha x+c)}$

where

${\displaystyle \alpha x={\frac {B}{A}}}$
${\displaystyle A=e^{c}}$

or

${\displaystyle f(x)=Ae^{(}-\alpha x)}$