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## 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

#### Step-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
 $I.F. = {e^{\int {\tan xdx} }} = {e^{\ln (\sec x)}} = \sec x$

#### Step-2

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 :
 $\dfrac{d}{{dx}}(y\sec x) = 1$ $\Rightarrow d\left( {y\sec x} \right) = dx$

#### Step-3

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