In: Math
Topic: Homogeneous equations:
Create a powerpoint and present one of these with explanations and at least 3 examples.
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:
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 =
c1em1x+c2em2x.
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 =
c1em1x+c2em2x
y =
c1ex+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 =
c1em1x+c2em2x
y =
c1ex+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' =
c1ex−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 =
2ex−e−3x
.