# The Actual Way to Solve Linear Differential Equation

Differential Equation / Saturday, September 8th, 2018

In this article we are going to learn all about Linear Differential Equation and how to solve it.

# Linear Differential Equation Definition

A differential equation that can be put in the form
$\frac{dy}{dx}+Py=Q$
where P and Q are functions of x alone, or constant, is called linear differential equation of first order in y.

OR,

$\frac{dx}{dy}+Px=Q$
where P and Q are functions of y alone, or constant, is called linear differential equation of first order in x.
The dependent variable y as well as its derivative in such equations occurs in the first degree only and not as higher powers or products.

## Working Rule for Solving a Linear Differential Equation

It will be convenient for the students to follow the following four steps to solve a linear differential equation.
Step 1. The equation should be put in standard form to identify P and Q.
Step 2. ʃPdx or ʃPdy is to be calculated.
Then the integrating factor e ʃPdx or e ʃPdy is to be determined after proper simplification.
Step 3. The given equation is to be multiplied by this integrating factor.
Step 4. Finally, both sides of the resulting equation is to be integrated, remembering that the integral of the left hand side is always y times (or, x times) the integrating factor.

 Example 01

$Solve:\frac{dy}{dx}+y=x$
Solution:
$Given,\frac{dy}{dx}+y=x$
It is a linear equation in y.
Comparing the given equation with
$\frac{dy}{dx}+Py=Q$
$We-have:P=1,and,Q=x$
$\therefore I.F.={{e}^{\int{Pdx}}}={{e}^{\int{dx}}}={{e}^{x}}$
Multiplying both side of given equation by I.F. and integrating we get,
$y{{e}^{\int{Pdx}}}=\int{Q.{{e}^{\int{Pdx}}}dx+c}$
$\Rightarrow y{{e}^{x}}=\int{x{{e}^{x}}dx+c}$
$\Rightarrow y{{e}^{x}}=x\int{{{e}^{x}}dx-\int{\left\{ \frac{d}{dx}\left( x \right)\int{{{e}^{x}}dx} \right\}}dx+c}$
$\Rightarrow y{{e}^{x}}=x{{e}^{x}}-\int{{{e}^{x}}dx+c}$
$\Rightarrow y{{e}^{x}}=x{{e}^{x}}-{{e}^{x}}+c$
$\therefore y=x-1+c{{e}^{-x}}$
is the required solution.

 Example 02

$Solve:\frac{dy}{dx}+y\cot x=2\cos x$
Solution:
$Given,\frac{dy}{dx}+y\cot x=2\cos x$
It is a linear equation in y.
$\therefore I.F.={{e}^{\int{\cot xdx}}}={{e}^{\log \sin x}}=\sin x$
Multiplying both side of given equation by I.F. and integrating we get,
$y\sin x=\int{2\cos x.\sin xdx+c}$
$\Rightarrow y\sin x=\int{\sin 2xdx+c}$
$\therefore y\sin x=-\frac{\cos 2x}{2}+c$
is the required solution.

 Example 03

$Solve:{{\cos }^{2}}x\frac{dy}{dx}+y=\tan x$
Solution:
$Given,{{\cos }^{2}}x\frac{dy}{dx}+y=\tan x$
$\Rightarrow \frac{dy}{dx}+y{{\sec }^{2}}x=\tan x{{\sec }^{2}}x$
It is a linear equation in y.
$\therefore I.F.={{e}^{\int{{{\sec }^{2}}xdx}}}={{e}^{\tan x}}$
Multiplying both side of given equation by I.F. and integrating we get,
$y{{e}^{\tan x}}=\int{\tan x{{\sec }^{2}}x{{e}^{\tan x}}}dx+c$
$Put,\tan x=z\Rightarrow {{\sec }^{2}}xdx=dz$
$\therefore y{{e}^{\tan x}}=\int{z{{e}^{z}}dz+c}$
$\Rightarrow y{{e}^{\tan x}}=z\int{{{e}^{z}}dz-\int{\left\{ \frac{d}{dz}\left( z \right)\int{{{e}^{z}}dz} \right\}}dz+c}$
$\Rightarrow y{{e}^{\tan x}}=z{{e}^{z}}-\int{{{e}^{z}}dz+c}$
$\Rightarrow y{{e}^{\tan x}}=z{{e}^{z}}-{{e}^{z}}+c$
$\Rightarrow y{{e}^{\tan x}}=\left( \tan x-1 \right){{e}^{\tan x}}+c$
is the required solution.

 Example 04

$Solve:\frac{dy}{dx}+\frac{y}{x}=\sin {{x}^{2}}$
Solution:
$Given,\frac{dy}{dx}+\frac{y}{x}=\sin {{x}^{2}}$
It is a linear equation in y.
$\therefore I.F.={{e}^{\int{\frac{1}{x}dx}}}={{e}^{\log x}}=x$
Multiplying both side of given equation by I.F. and integrating we get,
$yx=\int{x}\sin {{x}^{2}}dx+c$
$Put,{{x}^{2}}=z\Rightarrow xdx=\frac{1}{2}dz$
$\Rightarrow yx=\frac{1}{2}\int{\sin zdz+c}$
$\Rightarrow yx=-\frac{1}{2}\cos z+c$
$\Rightarrow yx=-\frac{1}{2}\cos {{x}^{2}}+c$
is the required solution.

 Example 05

$Solve:\cos x\frac{dy}{dx}+y\sin x=1$
Solution:
$Given,\cos x\frac{dy}{dx}+y\sin x=1$
$\Rightarrow \frac{dy}{dx}+y\tan x=\sec x$
It is a linear equation in y.
$\therefore I.F.={{e}^{\int{\tan xdx}}}={{e}^{\log \sec x}}=\sec x$
Multiplying both side of given equation by I.F. and integrating we get,
$y\sec x=\int{{{\sec }^{2}}x}dx+c$
$\therefore y\sec x=\tan x+c$
is the required solution.

 Example 06

$Solve:\frac{dy}{dx}+y\sec x=\tan x$
Solution:
$Given,\frac{dy}{dx}+y\sec x=\tan x$
It is a linear equation in y.
$\therefore I.F.={{e}^{\int{\sec xdx}}}={{e}^{\log \left( \sec x+\tan x \right)}}=\sec x+\tan x$
Multiplying both side of given equation by I.F. and integrating we get,
$y\left( \sec x+\tan x \right)=\int{\tan x\left( \sec x+\tan x \right)}dx+c$
$\Rightarrow y\left( \sec x+\tan x \right)=\int{\sec x\tan xdx+\int{{{\tan }^{2}}xdx+c}}$
$\Rightarrow y\left( \sec x+\tan x \right)=\sec x+\int{\left( {{\sec }^{2}}x-1 \right)}dx+c$
$\Rightarrow y\left( \sec x+\tan x \right)=\sec x+\tan x-x+c$
$\therefore \left( \sec x+\tan x \right)\left( y-1 \right)+x=c$
is the required solution.

 Example 07

$Solve:x\frac{dy}{dx}={{x}^{2}}+3y$
Solution:
$Given,x\frac{dy}{dx}={{x}^{2}}+3y$
$\Rightarrow \frac{dy}{dx}=x+\frac{3}{x}y$
$\Rightarrow \frac{dy}{dx}-\frac{3}{x}y=x$
It is a linear equation in y.
$\therefore I.F.={{e}^{\int{-\frac{3}{x}dx}}}={{e}^{-3\log x}}={{e}^{\log {{x}^{-3}}}}=\frac{1}{{{x}^{3}}}$
Multiplying both side of given equation by I.F. and integrating we get,
$y.\frac{1}{{{x}^{3}}}=\int{x.\frac{1}{{{x}^{3}}}dx+c}$
$\Rightarrow \frac{y}{{{x}^{3}}}=\int{\frac{1}{{{x}^{2}}}dx+c}$
$\Rightarrow \frac{y}{{{x}^{3}}}=-\frac{1}{x}+c$
$\therefore y+{{x}^{2}}=c{{x}^{3}}$
is the required solution.

 Example 08

$Solve:\frac{dy}{dx}+\left( \frac{1}{1+{{x}^{2}}} \right)y=\frac{{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}}$
Solution:
$Given,\frac{dy}{dx}+\left( \frac{1}{1+{{x}^{2}}} \right)y=\frac{{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}}$
It is a linear equation in y.
$\therefore I.F.={{e}^{\int{\frac{1}{1+{{x}^{2}}}dx}}}={{e}^{{{\tan }^{-1}}x}}$
Multiplying both side of given equation by I.F. and integrating we get,
$y{{e}^{{{\tan }^{-1}}x}}=\int{{{e}^{{{\tan }^{-1}}x}}}.\frac{{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}}dx+c$
$Put,{{\tan }^{-1}}x=z\Rightarrow \frac{1}{1+{{x}^{2}}}dx=dz$
$\Rightarrow y{{e}^{{{\tan }^{-1}}x}}=\int{{{e}^{2z}}}dz+c$
$\Rightarrow y{{e}^{{{\tan }^{-1}}x}}=\frac{{{e}^{2z}}}{2}+c$
$\Rightarrow y{{e}^{{{\tan }^{-1}}x}}=\frac{{{e}^{2{{\tan }^{-1}}x}}}{2}+c$
is the required solution.

 Example 09

$Solve:\left( 1+{{x}^{2}} \right)\frac{dy}{dx}+y={{\tan }^{-1}}x$
Solution:
$Given,\left( 1+{{x}^{2}} \right)\frac{dy}{dx}+y={{\tan }^{-1}}x$
$\Rightarrow \frac{dy}{dx}+\frac{y}{1+{{x}^{2}}}=\frac{{{\tan }^{-1}}x}{1+{{x}^{2}}}$
It is a linear equation in y.
$\therefore I.F.={{e}^{\int{\frac{1}{1+{{x}^{2}}}dx}}}={{e}^{{{\tan }^{-1}}x}}$
Multiplying both side of given equation by I.F. and integrating we get,
$y{{e}^{{{\tan }^{-1}}x}}=\int{{{e}^{{{\tan }^{-1}}x}}.}\frac{{{\tan }^{-1}}x}{1+{{x}^{2}}}dx+c$
$Put,{{\tan }^{-1}}x=z\Rightarrow \frac{1}{1+{{x}^{2}}}dx=dz$
$\Rightarrow y{{e}^{{{\tan }^{-1}}x}}=\int{z{{e}^{z}}dz+c}$
$\Rightarrow y{{e}^{{{\tan }^{-1}}x}}=\left( z-1 \right){{e}^{z}}+c$
$\therefore y{{e}^{{{\tan }^{-1}}x}}=\left( {{\tan }^{-1}}x-1 \right){{e}^{{{\tan }^{-1}}x}}+c$
$\therefore y=\left( {{\tan }^{-1}}x-1 \right)+c{{e}^{-{{\tan }^{-1}}x}}$
is the required solution.

 Example 10

$Solve:1+{{y}^{2}}+\left( x-{{e}^{-{{\tan }^{-1}}y}} \right)\frac{dy}{dx}=0$
Solution:
$Given,1+{{y}^{2}}+\left( x-{{e}^{-{{\tan }^{-1}}y}} \right)\frac{dy}{dx}=0$
The given equation can be written as
$1+{{y}^{2}}=-\left( x-{{e}^{-{{\tan }^{-1}}y}} \right)\frac{dy}{dx}$
$\Rightarrow \left( 1+{{y}^{2}} \right)\frac{dx}{dy}=-x+{{e}^{-{{\tan }^{-1}}y}}$
$\Rightarrow \frac{dx}{dy}+\frac{x}{1+{{y}^{2}}}=\frac{{{e}^{-{{\tan }^{-1}}y}}}{1+{{y}^{2}}}$
It is a linear equation in x.
$\therefore I.F.={{e}^{\int{\frac{1}{1+{{y}^{2}}}dy}}}={{e}^{{{\tan }^{-1}}y}}$
Multiplying both side of given equation by I.F. and integrating we get,
$x{{e}^{{{\tan }^{-1}}y}}=\int{{{e}^{{{\tan }^{-1}}y}}}.\frac{{{e}^{-{{\tan }^{-1}}y}}}{1+{{y}^{2}}}dy+c$
$\Rightarrow x{{e}^{{{\tan }^{-1}}y}}=\int{\frac{1}{1+{{y}^{2}}}dy}+c$
$\therefore x{{e}^{{{\tan }^{-1}}y}}={{\tan }^{-1}}y+c$
is the required solution.

 Example 11

Find a solution of the equation
$\frac{dy}{dx}+y\cot x=5{{e}^{\cos x}}$
Satisfying the initial conditions x= π/2, y=-4
Solution:
The given equation is
$\frac{dy}{dx}+y\cot x=5{{e}^{\cos x}}$
It is a linear equation.
$\therefore I.F.={{e}^{\int{\cot xdy}}}={{e}^{\log \sin x}}=\sin x$
Multiplying both side of given equation by I.F. and integrating we get,
$y\sin x=5\int{{{e}^{\cos x}}.\sin xdx+c}$
$\Rightarrow y\sin x=-5\int{{{e}^{\cos x}}d\left( \cos x \right)+c}$
$\Rightarrow y\sin x=-5{{e}^{\cos x}}+c$
Now initially, x= π/2, y=-4
$\therefore -4.\sin \frac{\pi }{2}=-5{{e}^{\cos \frac{\pi }{2}}}+c$
$\Rightarrow -4=-5+c\Rightarrow c=1$
Therefore the required solution is
$y\sin x=-5{{e}^{\cos x}}+1$