Calculus Course/Differential Equations/1st Order Differential Equations

1st Order Differential Equations

A 1st order differential equation has the form shown below

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

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

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

$\alpha ={\frac {B}{A}}$

Proof

The above equation can be rewritten as

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

Then

${\frac {df(x)}{dx}}=-{\frac {B}{A}}f(x)$

$\int {\frac {df(x)}{f(x)}}=-{\frac {B}{A}}\int dx$

$Lnf(x)=-{\frac {B}{A}}x+C$

$f(x)=e^{(}-{\frac {B}{A}}x+C)$

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

Summary

First ordered differential equation of the form

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

has a exponential root of the form

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

where

$\alpha x={\frac {B}{A}}$

$A=e^{c}$

or

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