Question

Topic: Homogeneous equations:

Create a powerpoint and present one of these with explanations and at least 3 examples. 

Answer 1

Homogeneous Equations:

Before starting Homogeneous Equations let us know it's background.One of the most important branches of maths is calculus. It plays an important role in physics, engineering, chemistry and in many more scientific fields.

Calculus can be divided into two main parts:
one is differential calculus and the other is integral calculus. As the name itself suggests that differential calculus deals with differentiation; while integral calculus is concerned with integration. The study of differential equations is a major part of differential calculus.

A differential equation may be defined as an equation containing a function as well its derivatives. There are different classifications of the differential equations :

On the Basis of Type of Derivatives:
i) Ordinary Differential Equations:
              The equations containing derivatives of the function which possesses only one independent variable with respect to which differentiation is done.

ii) Partial Differential Equations:
              The equations containing derivatives of the function which possesses more than one independent variables. Thus, the equation involves partial derivatives.

On the Basis of Type of Linearity:
i) Linear Differential Equations:
             The differential equation with the derivatives to be in the power one.

ii) Non-linear Differential Equations:
            The equation that are not categorized as linear are known as nonlinear equations.

On the Basis of Homogeneity:
i) Homogeneous Differential Equations----------------------------------------->>>>>>This is our concerned Topic
       The equations having right hand side equal to zero.

ii) Non-homogeneous Differential Equations
          The equations which right hand side not equal to zero, but a function of the independent variable.
Let us go ahead in this page and learn more about homogeneous equations.

Definition:   A linear differential equation of the form:

1. f₀₍x) y+f₁(x) y′+f₂(x) y′′+...+f_n(x) y_n=g(x)

is known as a Homogenous equation if and only if g(x) = 0

f g(x) ≠ 0, it is said to be a non-homogeneous equation.

We can also define a first-order linear homogeneous differential equation to be of the following form :
dy/dx+y f(x)= 0

1.The examples are first-order homogeneous equations are:

a) dy/dx - 5y = 0

b) 4dy/dx - y = 0

How to Solve:

First-Order Homogeneous Equations:
These equations are usually separable equations, so we separate the variables. We mean to say that the equation should be expressed in the form f(y) dy = g(x) dx and then this equation is to be integrated both the sides. In this way, we easily reach the solution.

Example 1: Solve the following equation:

dy/dx = 3xy

Solution: dy/dx = 3xy

dy/y = 3xdx

Integrating on both sides, we get

∫ dy/y = 3 ∫x dx

log y = 3 x²/2 + c

log y = 3/2x² + c

Second-Degree Homogeneous Equation:

The general form of a second-order homogeneous equation is given below:

a₀y+a₁y′+a₂y′′=0

Where, a2≠0

and a0, a1, a2 are constant. So this is known as second-order homogeneous equation

with constant coefficients.

It can also be written in the following form

a₀y+ a₁ dy/dx + a₂ (dy/dx)² = 0

Sometimes, the notation D is used for dy/dx
i.e.
dⁿy/dxⁿ = Dⁿ

The examples are second-order homogeneous equations are:
a) y'' - 5y' + 6 = 0
b) 4D² - 3D - 1 = 0

How to Solve:
Second-Order Homogeneous Equations:
There is a special technique for solving these equations. One should follow the instruction written below :
Step 1: Let the given equation be "a D² + b D + cy = 0".First, we shall the characteristic equation which is

am² + bm + c = 0.

Step 2: Now, suppose m1 and m2 be two distinct real numbers that are the solution of given equation. We shall find out the solution of above characteristic equation just as a quadratic equation.

Step 3: The general solution of given differential equation would be y = c1e^m1x+c2e^m2x.

Step 4: In case initial values are given, we may even find the values of D = dy/dx and D² = d²y/dx² from the equation of y and substitute these values in the given differential equations.

Example 1: Find the general solution of the following second-order differential equation:

D2+D−2y=0 where D = dydx

Solution: D2+D−2y=0

The corresponding characteristic equation is:

m2 + m - 2 = 0

m2 + 2m - m - 2 = 0

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

m = 1, -2

i.e. m1 = 1, m2 = -2

Thus, general solution would be

y = c₁e^m1x+c₂e^m2x

y = c₁e^x+c₂e^−2x

Example 3: Solve the differential equation y″ + 2y′ - 3y = 0. Calculate for initial value problem at y(0) = 1 and y'(0) = 5

Solution: y″ + 2y′ - 3y = 0

Its characteristic equation is:

m2 + 2m - 3 = 0

m2 + 3m - m - 3 = 0

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

m = 1, -3

i.e. m1 = 1, m2 = -3

Thus, general solution would be

y = c₁e^m1x+c₂e^m2x

y = c₁e^x+c₂e^−3x_______(1)

Now, For IVP

y(0) = 1 and y'(0) = 5

Substituting x = 0 and y = 1 in equation (1)

c1+c2=1 ________(2)

Differentiating equation (1), we get

y' = c₁e^x−3c₂e^−3x

Substituting x = 0 and y' = 5 in above equation

c₁−3c₂=5 ________(3)

Solving equation (2) and (3), we get

c1 = 2 and c2 = -1

Plugging these values in equation (1), we obtain

y = 2e^x−e^−3x   

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