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

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

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

Proof

The above equation can be rewritten as

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

Then

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

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

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

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

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

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

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

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