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.
     Example: 1   

Solve the DE \dfrac{{dy}}{{dx}} + y\tan x = \cos x
Solution: 1


Comparing this DE with the standard form of the linear DE \dfrac{{dy}}{{dx}} + Py = Q, we see that
P(x) = \tan x,  Q(x) = \cos x
Thus, the I.F. is
I.F. = {e^{\int {\tan xdx} }} = {e^{\ln (\sec x)}} = \sec x


Multiplying by \sec x on both sides of the given DE, we obtain
\sec x\dfrac{{dy}}{{dx}} + y\tan x\sec x = 1
The left hand side is an exact differential :
\dfrac{d}{{dx}}(y\sec x) = 1
 \Rightarrow  d\left( {y\sec x} \right) = dx


Integrating both sides, we obtain the solution to our DE as
y\sec x = x + C
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