ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 11- First Order Linear DEs

Tuesday, 5 August 2014

CHAPTER 11- First Order Linear DEs

We now come to a very important class of DEs : first-order linear DEs, their importance arising from the fact that many natural phenomena can be described using such DEs.

First order linear DEs take the form

$\dfrac{{dy}}{{dx}} + P(x)y = Q(x)$

where $P$ and $Q$ are functions of $x$ alone.

To solve such DEs, the method followed is as described below :

We multiply both sides of the DE by a quantity called the integrating factor (I.F.) where

$I.F. = {e^{\int {Pdx} }}$

Why this is chosen as the I.F. will soon become clear when we see what the I. F. actually does :

${e^{\int {Pdx} }}\left( {\dfrac{{dy}}{{dx}} + Py} \right) = {e^{\int {Pdx} }} \cdot Q$

The left hand side now becomes exact, in the sense that it can be expressed as the exact differential of some expression :

${e^{\int {Pdx} }}\left( {\dfrac{{dy}}{{dx}} + Py} \right) = \dfrac{d}{{dx}}\left( {y{e^{\int {Pdx} }}} \right)$

Now our DE becomes

$\dfrac{d}{{dx}}\left( {y{e^{\int {Pdx} }}} \right) = Q \cdot {e^{\int {Pdx} }}$

This can now easily be integrated to yield the required general solution:

$y{e^{\int {Pdx} }} = \int {\left( {Q{e^{\int {Pdx} }}} \right)dx + C}$

You are urged to re-read this discussion until you fully understand its significance. In particular, you must understand why multiplying the DE by the I. F. ${e^{\int {Pdx} }}$ on both sides reduces its left hand side to an exact differential.