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Homogeneous Linear Differential Equations

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An equation is a statement that equates two mathematical expressions. A differential equation consists of a function and its derivatives.  A differential equation will consist of dependent and independent variables. If we have y = f(x), then x is an independent variable while y is an dependent variable.
 
A differential equation can be classified based on the following parameters:

1) Type of derivatives: Ordinary and Partial
An ordinary differential equation has only one independent variable whereas partial differential equation has more than one independent variables. y' + xy = 0 is an ordinary differential equation whereas y' + xz = 0 is a partial differential equation.

2) Type of linearity: Linear and non-linear
Linear differential equations have derivatives of power one which is not true for non-linear differential equations. Non-linear equations have derivatives of higher power. We have y'' + y = 0 as a linear differential equation whereas y''2 + y = 0 as non-linear differential equation.

3) Type of homogeneity: Homogenous and non-homogenous
Homogenous differential equation will have the right side of the equation equal to zero. The non-homogenous equation will have a function of the independent variable on the right hand side. The equation y'' + y = 2x + 3 is a non-homogenous differential equation but y'' + y = 0 is a homogenous differential equation.

Definition

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Homogenous linear differential equations, as the name suggests, have the derivatives of power one and the right hand side equals to zero.

A homogenous linear differential equation will be in the given form:

$f_{n}y^{n}+...+f_{1}y^{1}+f_{0}y^{0}=0$ where yn is the n-th derivative.

Let the solution be in the form y(t) = $e^{rt}$ . Then, $y^1$ = $re^{rt}$ , $y^2$ = r2ert, and so on.

Putting it in the homogenous linear equation we get,

$e^{rt}[a_{n}r^{n}+a_{n-1}r^{n-1}+...+a_{1}r^{1}+a_{0}]= 0$

which results in $[a_{n}r^{n}+a_{n-1}r^{n-1}+...+a_{1}r^{1}+a_{0}]= 0$.

This is known as the characteristic equation. For an nth degree equation, there will be n roots.

With Constant Coefficients  

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A homogenous linear differential equation where all the coefficients are constant terms is known as a homogenous linear differential equation with constant coefficient.
To summarize, such equation will have following attributes:

1)  Derivatives of the function and the function

2) All derivatives have power of one

3) Right hand side of the equation is zero

4) All coefficients are constants

Let us take a second order linear differential equation with constant coefficients. It will be of the form ay'' + by' + c = 0. The characteristic equation will be $ar^2$+ br + c = 0, by putting y = $e^{rt}$ as the solution. As the power is 2, there will be two roots, $r^1$ and $r^2$ .

1)  If $r^1$ and $r_2$ are two real and distinct roots then the general solution will be

y = $c_1 e^{r_1 x}$ + $c_2 e^{r_2x}$

2) If $r_1$ = $r_2$ , then the general solution will be 

y = $c_1 e^{r_1x}$ + $xc_2e^{r_1x}$ 

3) If $r_1$ and $r_2$ are complex and conjugate, then the general solution will be

 $y = c_{1}e^{\alpha x}cos\beta x + c_{2}e^{\alpha x}sin\beta x$

When initial conditions are given, they are used to get the value of the constants and finally the actual solution. For example, if the roots are coming 2, 1 and initial conditions are given as y(0) = 1, y'(0) = 2. As roots are real and distinct, the general solution will be

y = $c_1 e^{2x}$ + $c_2^{ex}$ 

y(0) = $c_1$ + $c_2$ = 1

y'(0) = 2$c_1$ + $c_2$ = 2

Solving we get, $c_1$ = 1; $c_2$ = 0.

Hence, actual solution becomes y = $e^{2x}$.

Examples

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Example 1: Find the general solution of the given equation.

y'' - 2y' + y = 0

Solution: Taking y = $e^{rt}$ we get,

$r^2$ - 2r + 1 = 0. Solving the quadratic equation we get only one root, r = 1.

Hence, the general solution will be y =  $c_1e^x$ + $xc_2e^x$. 

Example 2: Find the general solution of the given equation.

y'' + 2y' - 3y = 0

Solution: Taking solution as y = $e^{rt}$ we get the characteristic equation as

$r^2$  + 2r -3 = 0

$r^2$  + 3r - r - 3 = 0

r(r + 3) - 1(r + 3) = 0

r = 1, -3

As both roots are real and distinct the general solution will be,

y =  $c_1e^x$ + $c_2e^{-3x}$ 

Example 3: Find the solution of the given equation if the initial conditions are y(0) = 2,

y'(0) = -1.

y'' + y' - 2 = 0

Solution: Taking solution as y = $e^{rt}$ we get the characteristic equation as:

$r^2$ + r - 2 = 0

$r^2$ + 2r - r - 2 = 0

(r + 2)(r - 1) = 0

r = -2, 1

The general solution will be y =  $c_1e^x$ + $c_2e^{-3x}$ 

Now, using the initial conditions we get

y(0) = $c_1$ + $c^2$ = 2

y'(0) = -2$c_1$ + $c^2$= -1

Solving we have $c_1$ =  $c^2$ = 1.

The actual solution is y = $e^{-2x}$+ $e^x$ .