Download Pertemuan VI - Binus Repository

PERTEMUAN VI.
Penyelesaian Khusus (Particular Solution )
Nonhomogeneous Linear Equations
Consider the nonhomogeneous linear equation
We have seen that the general solution is given by
,
where
is a particular solution and
is the general solution of the associated homogeneous equation. We will not
discuss the case of non-constant coefficients. Therefore, we will restrict
ourself only to the following type of equation:
Using the previous section, we will discuss how to find the general solution of
the associated homogeneous equation
Therefore, the only remaining obstacle is to find a particular solution to (NH).
In the second order differential equations case, we learned the two methods:
Undetermined Coefficients Method and the Variation of Parameters . These
VI-1
two methods are still valid in the general case, but the second one is very hard
to carry.
Undetermined Coefficients Method
Example: Find a particular solution of
Solution: Let us follow these steps:
(1) Characteristic equation
We have the factorization
.
Therefore, the roots are 0, 2, -2 and they are all simple.
(2) We have to split the equation into the following two equations:
;
(3) The particular solution to the equation (1):
(3.1) We have
which is a simple root. Then s = 1;
(3.2) The guessed form for the particular solution is
,
where A and B are to be determined. We will omit the detail of the
calculations. We get A = -1/8 and B = 0. Therefore, we have
VI-2
;
(4) The particular solution to the equation (2):
(4.1) We have
which is not a root. Then s = 0;
(4.2) The guessed form for the particular solution is
where A and B are to be determined. We will omit the detail of the
calculations. We get A = 0 and B=-3/5. Therefore, we have
;
(5) The particular solution to the original equation is given by
(4) The particular solution to the equation (2):
(4.1) We have
which is not a root. Then s = 0;
(4.2) The guessed form for the particular solution is
where A and B are to be determined. We will omit the detail of the
calculations. We get A = 0 and B=-3/5. Therefore, we have
;
(5) The particular solution to the original equation is given by
VI-3
Method of Variation of Parameters
This method has no prior conditions to be satisfied. Therefore, it may sound
more general than the previous method. We will see that this method depends
on integration while the previous one is purely algebraic which, for some at
least, is an advantage.
Consider the equation
In order to use the method of variation of parameters we need to know that
is a set of fundamental solutions of the associated homogeneous equation y'' +
p(x)y' + q(x)y = 0. We know that, in this case, the general solution of the
associated homogeneous equation is
. The idea behind the method of variation of parameters is to look for a
particular solution such as
where and are functions. From this, the method got its name.
The functions and are solutions to the system
,
which implies
,
VI-4
where
is the wronskian of
and
. Therefore, we have
Summary: Let us summarize the steps to follow in applying this method:
(1) Find
a set of fundamental solutions of the associated
homogeneous equation
y'' + p(x)y' + q(x)y = 0.
(2) Write down the form of the particular solution
;
(3) Write down the system
;
(4) Solve it. That is, find
(5) Plug
and
and
;
into the equation giving the particular solution.
Example: Find the particular solution to
Solution: Let us follow the steps:
(1) A set of fundamental solutions of the equation y'' + y = 0 is
;
(2) The particular solution is given as
VI-5
(3) We have the system
;
(4) We solve for
and
, and get
Using techniques of integration, we get
(5) The particular solution is:
or
Remark: Note that since the equation is linear, we may still split if necessary.
For example, we may split the equation
,
into the two equations
then, find the particular solutions
for (1) and
particular solution for the original equation by
for (2), to generate a
VI-6
Metode Singkat
( Dn + bn-1 Dn-1 + b0D ) y = g(x)
F(D) = g(x)
1. Bila
g(x) = e
Maka
yp =
ax
1
g ( x)
D  bn 1 D n 1 , , , b0 D
n
1
1 ax
e ax 
e ,
F ( D)
F (a)
=
1
e ax ,
F (a)
Bila
F(a)
= 0 maka
yp = x
Bila
F’(a)
= 0 maka
yp = x 2
2. Bila
1
e ax ,
F " (a)
g(x) = xn
Maka
=
'
yp =
1
g ( x)
D  bn 1 D n 1 , , , b0 D
n
1
diuraikan
F ( D)
1
xn
F ( D)
3. Bila g(x) = sin ax
Maka
=
yp =
atau cos ax
1
g ( x)
D  bn 1 D n 1 , , , b0 D
n
1
sin ax
F (D 2 )
VI-7
=
1
sin ax
F (a 2 )
4. Bila g(x) = eax φ(x)
Maka
yp = eax
1
 ( x) =
F ( D  a)
1
 ( x)
F ( D  a)
e-a x ∫e-ax φ (x) dx
Example:
Find a particular solution to the equation
Solution: Let us follow these steps:
(1) First, we notice that the conditions are satisfied to invoke the method
of undetermined coefficients.
(2) We split the equation into the following three equations :
(2) The root of the characteristic equation
and r = 4.
are r = -1
Particular solution to Equation (1):
Since
, and
, then
, which is not one of the
roots. Then s = 0.
The particular solution is given as
VI-8
If we plug it into the equation (1), we get
,
which implies A = -1/2, that is,
Particular solution to Equation (2):
Since
, and β = 1, then α + βi = i , which is not one of the roots.
Then s = 0.
The particular solution is given as
If we plug it into the equation (2), we get
,
which implies
Particular solution to Equation (3):
Since α = -1, and β = 0 , then α + βi = -1 which is one of the roots.
Then s = 1.
VI-9
The particular solution is given as
If we plug it into the equation (3), we get
,
which implies
, that is
(5) A particular solution to the original equation is
Example 3. Apply the First Shift Theorem to the following differential
equations to determine the particular integral for each:
a.
b.
c.
Example 4. Give the general solutions to the following differential equations:
a. y + 4y + 3y = e-2x;
b. y + 4y + 3y = 5 cos(2x);
VI-10
Example 5. Solve the following differential equation: y + 4y + 4y = 24
sin(2x), given that y(0) = 1 and y'(0) = 2.
Solucia
The menu of this page is in simple mode. There is a menu in expert mode for
more
options.
The expert menu enable to get some animated curves of solutions.
y' + 2*y = x+4
Show

Resolution of the differential equation :
General solution.
Cauchy's solution.
In this case , give the initial value(s) below , separate with comma :
I.C. :

.
Curves representations:
initial values of "I.C.") ,
in the interval from
general solutions
to
In option , amplitude : from
one solution (with
.
to
.
To know more , click here .
Solucia
Attention,a differential equation of order 1 means 1 initial condition.
But you have gave 0 of those.
VI-11

The general solution is :

To obtain a particular solution of the equation, It's missing the initial
value.

Here is the evolution of the curve according to the missing value :
TERIMA KASIH
VI-12