differential eqations
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1
Differential Equations
1. Introduction
An equation involving one or more derivatives of an unknown function is called a differential
equation. Differential equations arise in many physical phenomena and mathematical analysis of
any engineering problems.
A mathematician is interested in proving that a given differential equation possesses a solution;
hence one can obtain the solution and deduce a few properties of that solution. A physicist or an
engineer on the other hand is usually interested in the specific expression of the solution. The
usual compromise is to find the solution.
1.1 Fundamental definitions
An ordinary differential equation is an equation which involves derivatives of an unknown
function y of a single variable x.
Ordinary differential equations (ODEs) arise in many different contexts throughout mathematics
and science (social and natural) one way or another, because when describing changes
mathematically, the most accurate way is differentials and derivatives (related, though not quite
the same). Since various differentials, derivatives, and functions become inevitably related to
each other via equations, a differential equation is the result, describing dynamical phenomena,
evolution, and variation. Often, quantities are defined as the rate of change of other quantities
(time derivatives), or gradients of quantities, which is how they enter differential equations.
Specific mathematical fields include geometry and analytical mechanics. Scientific fields include
much of physics and astronomy (celestial mechanics), geology (weather modeling), chemistry
(reaction rates), biology (infectious diseases, genetic variation), ecology and population
modeling (population competition), economics (stock trends, interest rates and the market
equilibrium price changes). Many mathematicians have studied differential equations and
contributed to the field, including Newton, Leibniz, the Bernoulli family, Riccati, Clairaut,
d'Alembert, and Euler. A simple example is Newton's second law of motion — the relationship
between the displacement x and the time t of the object under the force F, which leads to the
differential equation.
Examples:
The order of a differential equation is the order of the highest derivative occurring in it.
The degree of a differential equation is the degree of the highest derivative exists in the
equation, provided the equation is free of radicals and terms with fractional degree.
The order and degree of the differential equations in the above examples are respectively
2
A partial differential equation is an equation which involves partial derivatives of unknown
functions of two or more independent variables.
Here, we consider only ordinary differential equations.
A differential equation is said to be linear if it is a linear function of the dependent variable and
its derivatives, i.e., it is of degree one in the dependent variable y and its derivatives.
The general linear differential equation of order n is of the form
Where b0, b1,…….. bn, and R(x) are functions of x alone.
A solution of a differential equation is a relation between the dependent and independent
variables, having derivative of order equal to order of the differential equation ,which is also free
of the derivatives and satisfying the given differential equation.
Example 1.1
The solution of is
Since .
This solution is a particular solution. The expression also satisfies the equation and it is
the general solution of this equation, for any arbitrary c.
Example 1.2
Consider the differential equation 2
23 2 0
d y dyy
dxdx .
Let y = e2x
, then 2
2 2
22 4 .x xdy d ye and e
dx dx
Hence 2
23 2 0
d y dyy
dxdx implying that y = e
2x is a solution of the given differential equation.
Also y = ex is a solution. These two solutions are particular solutions of the given equation. Note
that y = C1 e2x
+ C2 ex is also a solution , where C1and C2 are arbitrary constants. This solution is
called the general solution of the given differential equation, which is the linear combination of
all possible linearly independent solutions.
Note:
A differential equation together with an initial condition is called an Initial Value problem. The
initial condition is used to determine the value of the arbitrary constants in the general solution.
1.2 Formulation of differential equations by eliminating arbitrary constants
In practice, differential equations arise in many ways, one of which is useful in that it gives us a
feeling for the kinds of solutions to be expected. In this section we start with the relation
3
involving arbitrary constants, and, by elimination of those arbitrary constants obtain a differential
equation which is consistent with the original relation. In other words we will obtain a
differential equation for which the given relation is the general solution.
Methods for elimination of arbitrary constants vary the way in which the constants enter the
given relation. Since each differentiation yields a new relation, the number of derivatives that
needs to be used is same as that of the number of arbitrary constants to be eliminated. Thus in
eliminating arbitrary constants from a relation we obtain a differential equation that is
(i) Of order equal to the number of arbitrary constants in the equation.
(ii) Consistent with relation.
(iii)Free from arbitrary constants.
Example 1.2.1 Eliminate the arbitrary constants c1 and c2 from the relation
2 3
1 2 . (1)x xy c e c e
Since two constants are to be eliminated, obtain the two derivatives, 2 3
1 2
2 3
1 2
2 3 , (2)
4 9 . (3)
x x
x x
y c e c e
y c e c e
The elimination of 1 c from equations (2) and (3) yields
3
22 15 ;xy y c e
the elimination of 1 c from equations (1) and (2) yields
3
22 5 .xy y c e
Hence 2 3( 2 ),or 6 0.y y y y y y y
Another method for obtaining the differential equation in this example proceeds as follows. We
know from a theorem in elementary algebra that the equations (1), (2), and (3) considered as
equations in the two unknowns 1 2 and c c can have solutions only if
2 3
2 3
'' 2 3
2 3 0. (4)
4 9
x x
x x
x x
y e e
y e e
y e e
Since e-2x
and e3x
cannot be zero, equation (4) may be rewritten, with the factors e-2x
and e3x
removed, as
1 1
2 3 0
4 9
y
y
y
from which the differential equation 6 0y y y follows immediately.
4
This latter method has the advantage of making it easy to see that the elimination of the constants
1 2 , ,..., nc c c from a relation of the form 1 2m m
1 2e + e ... nm xx x
ny c c c e will always lead to a
differential equation 1
0 1 1... 0,
n n
nn n
d y d ya a a y
dx dx
in which the coefficients 0 1, ,..., na a a are constants. The
study of such differential equations will receive much of our attention.
Example 1.2.2 Eliminate the constant a from the equation 2 2 2( ) .x a y a
Direct differentiation of the relation yields 2( ) 2 0,x a yy from which .a x yy
Therefore, using the original equation, we find that 2 2 2 ,y x xyy which may be written in
the form 2 2( ) 2 0.x y dx xydy
Another method is based upon the isolation of an arbitrary constant.
The equation 2 2 2( )x a y a may be put in the form
2 2
2 .x y
ax
Differentiating both sides, we get 2 2
2
(2 2 ) ( )0,
x xdx ydy x y dx
x
or
2 2( ) 2 0,x y dx xydy as desired.
Example 1.2.3 Eliminate and B from the relation cos( ),- - - - - - - - (5)x B t in
which is a parameter (not to be eliminated).
First we obtain two derivatives of x with respect to t
22
2
sin( ), (6)
cos( ). (7)
dxB t
dt
d xB t
dt
On comparing equations (5) and (7) we get 2
2
20.
d xx
dt
Example 1.2.4 Eliminate c from the equation 2 4 0.cxy c x
Differentiating the given equation we get 2( ) 0.c y xy c
Since 0, ( )c c y xy and substitution into the original leads to the result
3 2 2( ) 4 0.x y x yy
Exercises 1.2.5
Form the differential equation by eliminating the arbitrary constants
1. 1 2cos sin ; a parameter.x c t c t
5
2. 2 4 .y ax
3. 2 2
1 2 .x xy x c e c e
4. 2 2 .x xy Ae Bxe
5. 1 2cos sin ; a and b are parameters.ax axy c e bx c e bx
1.3 Families of curves
A relation involving a parameter, as well as one or both the coordinates of a point in a plane,
represents a family of curves, one curve corresponding to each value of the parameter. For
instance, the equation 2 2 2( ) ( ) 2x c y c c may be interpreted as the equation of a family
of circles, each having its center on the line y = x and each passing through the origin.
If the constant c is treated as an arbitrary constant and eliminated, the result is called the
differential equation of the family represented by the equation. In this case, the elimination of c
is easily performed by isolating c, then differentiating throughout the equation with respect to x.
Thus, the given equation takes the form
2 2
2x y
cx y
and hence we can form the differential
equation.
Note that for a two parameter family of curves, the differential equation will be of order 2.
Example 1.3.1 Find the differential equation of the family of parabolas, having their vertices at
the origin and their foci on the y-axis.
From analytic geometry, we find the equation of this family of parabolas to be 2 4 .x ay Then
from F F
M and Nx y
then a may be eliminated by differentiation. Thus it follows
that 22 0.xydx x dy
Example 1.3.2 Find the differential equation of the family of circles having their centers on the
y-axis.
Since a member of the family of circles of this example may have its center anywhere on the y-
axis and its radius of any magnitude, we are dealing with the two-parameter family
2 2 2( ) .x y b r
Differentiating the given equation, we get
( ) 0,x y b y from which .x yy
by
6
Differentiating again,
2
2
[1 ( ) ] ( )0,
( )
y yy y y x yy
y
so the desired differential
equation is 3( ) 0.xy y y
Exercises 1.3.3
In each exercise, obtain the differential equation of the family of plane curves described and
sketch several representative members of the family.
1. Straight lines with slope and x-intercept equal.
2. Circles with fixed radius r and tangent to the x-axis.
3. Parabolas with axis parallel to the x-axis.
4. The cubics 2 2( ) cy x x a
with a fixed.
5. The cissoids
32 .
xy
a x
6. The strophoids (sec tan ).r a
1.4 Differential equations of order one and degree one
In our study we initially consider only first order and first degree differential equation. Such
equations can be written in the form
Before discussing some of the analytic techniques for finding solutions we shall state an
important theorem concerning to the uniqueness and existence of solution.
Consider
Let T denote the rectangular region defined by and , a region with
the point at its centre. Suppose that and are continuous functions of and in
Under the conditions imposed on above, an interval exists about , and
satisfies the properties.
a) A solution of equation (1) in
b) On the interval , satisfies
c) At ,
d) is unique in
satisfying the above conditions.
In other wards, the theorem states that if is sufficiently well behaved near the
point , then the differential equation ,
7
has a solution that passes through the point and that solution is unique near
We consider some first order differential equation in the following forms.
1.5 Variable Separable equations
If the differential equation M(x,y) dx + N(x,y) dy = 0 is simple enough that the variables can be
separated ,i.e., the equation can be rewritten as F(x) dx + G(y) dy = 0, where F is a function of
x alone and G is a function of y alone , then the solution can be obtained by direct integration.
Example 1.5.1
Integrating,
i.e. , is the required solution, where c is some arbitrary constant .
Example 1.5.2 Consider the equation,
Here the variables are not separated but it can be reduced to that form by dividing the equation
by Then
where the variables are separated.
The solution is
Remark: From the above example it is clear that an equation of the type
can be reduced to variable separable equation.
Example 1.5.3
Solution:
The solution is
Example 1.5.4
Solution:
8
Then given D.E becomes, 2 21 1 1 0x y dx x y dy
Example 1.5.5
Solution:
Then
Integrating,
Exercise 1.5.6
Solve
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
1.6 Differential equations with homogeneous coefficients:
Consider the differential equation of the form , where and are
homogeneous functions of the same degree in and .
9
To solve this equation, we note that being a homogenous equation of degree zero , is a
function of (y/x) only. Let . This suggests the substitution (y/x) = v or
then . Substituting in the given equation we get = g(v)
or = g(v) – v, which can be solved by separating the variables.
Example 1.6.1
Solution:
Put then
Example 1.6.2
Solution:
then
Integrating,
Remark: It is quite immaterial whether one uses or . However, it is sometimes
easier to solve by substituting for the variable whose differential has the simpler coefficient.
Example 1.6.3
Solution:
10
Integrating,
Exercise 1.6.4
Solve
1.
2.
3.
4.
5.
6.
7.
8.
1.7 Differential Equations with linear coefficients:
The equations of the form can be reduced to the homogenous form.
Case 1: When
Putting x= X + h and y = Y + k, where h, k are constants so that dx = dX , dy = dY
Choose h, k such that ah+bk+c=0 and 0 0a h b k c ab a b
and which is of homogenous coefficients in and can be solved.
Case 2: When i.e., , then equation can be reduced to variable-separable
Form by substituting and hence can be solved..
Example 1.7.1
Solve
Clearly,
11
put
Substituting,
Integrating both sides,
Finally, .
Example 1.7.2 Solve
Here
Put
i.e
i.e
Separating,
i.e Solution is
Exercise 1.7.3
Solve
1.
2.
3.
4.
5.
6.
7.
8.
9.
12
1.8 Exact Equations
These are equations of the type in which separation of variables
may not be possible. Suppose there exists a function such that is ,
then the solution is .
If the equation is exact, then by definition there exists a F such that
But from total derivate formula ,
Comparing the above equations, we get F F
M and Nx y
These two equations lead to and
Since ……………………(1)
Thus for an equation M dx + N dy = 0 to be exact it is necessary that the condition (1) is to be
satisfied. Now let us prove the converse of the above result. Let us assume that the condition (1)
holds. Let (x, y) be a function for which ( , )M x yx
. (i.e., the function is the result of
integrating M dx w.r.t. x alone holding y as a constant.
Then 2
;M
y x y
Hence from (1) ,
2 2 M N
x y y x y x
.
On integrating both sides of this equation w.r.t. x holding y fixed we get,
( ),N B yy
where B(y) is an arbitrary function of y.
Now define a function, ( , ) ( , ) ( ) .F x y x y B y dy
Then ( ) .F F
dF dx dy dx B y dy Mdx Ndyx y x y
Hence the given equation is exact. Thus , we have proved the following theorem.
Theorem: If M , N, M N
andy x
are all continuous functions of x and y , then a necessary and
sufficient condition for the differential equation M dx + N dy = 0 to be exact is that
Note : If the differential equation M dx + N dy = 0 is exact then the solution of the equation can
be obtained as follows. Let F be a function of x and y such that dF = M dx + N dy.
Then comparing with the equation F F
dF dx dyx y
we get F
x
= M (x, y).
13
On integrating w.r.t. x, holding y as a constant we get F(x, y) = ( , ) ( )M x y dx B y , where B(y)
is an arbitrary function of y alone. Now
( , ) ( ) ( , )F
M x y dx B y N x yy y
implies that ( )B y consists of terms in N(x, y) which
does not contain x.
Thus the solution function is given by F(x, y) = ( , ) ( )M x y dx B y
= ( , ) ( ( , ) )M x y dx termsin N x y not containg x dy
Example 1.8.1 Solve
M = 3x2y – 6x ……………..(1)
N = x2 +2y …………………..(2)
and .
Hence the equation is exact. Therefore the solution is given by
( , ) ( ( , ) )M x y dx termsin N x y not containg x dy = c
or,
Exercises 1.8.2:
Solve
1.
2.
3.
4.
5.
1.9 Linear Equations
Definition: A differential equation is said to be linear if the dependent variable and its
differential coefficient occur only in the first degree and not multiplied together.
A general linear differential equation is of the form
To solve, multiply both sides by P dx
e so that
The above equation is equivalent to
Integrating, we get as required solution.
If , then solution can be rewritten as
Example 1.9.1 Solve
Dividing throughout by ,
14
Thus the solution is .
Example 1.9.2 Solve
Which is linear equation in where
Thus the solution is
where
Exercises 1.9.3
Solve
1.
2.
3.
4.
5.
1.10. Bernoulli equation
The equation is reducible to the linear equation and is usually called
Bernoulli’s equation.
To solve, divide both sides by , so that .
Put . Equation reduces to which is linear in and
can be solved.
Example 1.10.1 Solve .
Dividing by , .
Put , so
15
i.e. , i.e which is linear in z.
I.F = = =
Solution is
i.e.
i.e.
Example 1.10.2 Solve
Dividing by ,
1 1
( )1
dxdy
y y x
i.e.
Exercise 1.10.3
Solve
1.
2.
3.
4.
5.
6.
7.
8. ,
1.11 Equations reducible to exact equations
Consider the equation which is not exact..…..(1)
Let , a function of and , be an integrating factor of (1). Then
must be exact.
Hence, must satisfy .
16
.
First let be a function of alone. Then and becomes .
Then we have or
If the left hand side of the above equation as function of alone, we have
Then the desired I.F is
By a similar argument, assuming is a function of alone, we get
Then an integrating factor is .
Using the above integrating factors, one can convert the equation to exact form and solve.
Example 1.11.1 Solve
and
Hence, .
.
Multiply to the above equation, we get
.
Solution is .
Example 1.11.2 Solve .
,
So
I.F =
Multiplying to the equation we get .
We get the solution .
Exercise 1.11.3 Solve
1.
2.
3.
4.
17
5.
6.
7.
8.
1.12 Integrating factors by inspection.
Here we are concerned with the equations that are simple enough to enable us to find the
integrating factors by inspection. The ability to do this depends largely upon recognition of
certain common exact differential and upon experience. Below are four exact differentials that
occur frequently:
Using these exact differentials it is possible to group the terms in given differential equation and
obtain the integrating factors.
Example 1.12.1 Solve .
Group the terms
Dividing by ,
Solution on integration, .
Example 1.12.2 Solve 2cos ( )dy y
x ydx x
2cos ( )y
xdy ydx dxx
2 2( ) cos ( )y y
d x dxx x
2
2sec ( ) ( )
y y dxd
x x x
1tan( )
yc
x x
Example 1.12.3 Solve 2 4 33 ( ) 0x ydx y x dy ,
Rewriting the given differential equation as 2 3 4(3 ) 0x ydx x dy y dy .
Which is nothing but 3 3 4( ) 0.yd x x dy y dy Dividing by y2, the equation becomes,
18
32 0
xd y dy
y
, which is exact. Therefore the solution is
3 3
3
x yC
y .
Exercise 1.12.4
Solve
1.
2.
3.
4.
5.
6.
7.
1.13 Higher Order Linear Differential Equations
The general linear differential equation of order n is an equation that can be written as
……………….(1)
If R(x) = 0, then the equation is called a homogenous linear differential equation otherwise it is
called non-homogeneous differential equation. The coefficient functions are
continuous on the interval I.
Consider .
If and are solutions of homogenous equation then is also a solution of
that equation.
In a similar manner, if y1 ,y2 ,…,yn are solutions of (1), then
……..(2)
is also a solution where are arbitrary constants.
The expression (2) is called linear combination of the functions .
For a homogeneous equation, the
general solution is of the form where are all possible
linearly independent solutions of the given equation. Where as for the nonhomogeneous equation
the general solution is of the
form c py y y , where cy is the general solution of the corresponding homogenous
equation and py is a particular
solution of the given equation , which does not contain any arbitrary constants.
1.3.1 An Existence and Uniqueness Theorem:
Given an order linear differential equation,
19
on an interval I,
suppose that is any number on the interval I and are n arbitrary constants.
Example 1.13.1 Find the unique solution of , .
Solution is .
Using we get .
1.13.2 Linear Independence of Solutions:
Given the functions if constants not all zero, exist such that
for all in , then the functions
are said to be linearly dependent on that interval. If no such relation exists, the functions are said
to be linearly independent.
1.13.3 The Wronskian of Solution:
To test whether n functions are linearly independent on an interval let us assume that
each of the functions is differentiable atleast ( times .
Then from the equation 1 1 2 2 ... n nc f c f c f = 0 , it follows by successive differentiation that
' ' '
1 1 2 2 ... n nc f c f c f = 0
''
1 1 2 2 ... n nc f c f c f = 0
… 1 1 1
1 1 2 2 ...n n n
n nc f c f c f = 0
For any fixed value of in , the nature of solutions of these will be determined by the
determinant
1 2
' ' '
1 2
1 1 1
1 2
( ) ( ) ... ( )
( ) ( ) ... ( )( )
. . . .
( ) ( ) ... ( )
n
n
n n n
n
f x f x f x
f x f x f xW x
f x f x f x
If for some on , then c1 = c2 =…=cn = 0 Hence, functions are linearly
independent on . The function is called Wronskian of the functions .
Example 1.13.2 Let 1 2 1 2, ,ax bx ax bxy e and y e then y ae y be .
( ) 0
ax bxa b x
ax bx
e eThen wronskian is given by W b a e
ae be
only if a=b.
Therefore for a≠b, 1 2ax bxy e and y e are linearly independent.
Example 1.13.3. Let 1 2 1 21 , 0, 1y and y x then y y
20
11
0 1
xThen wronskian is given by W ≠ 0. Therefore 1 21y and y x are linearly
independent.
Let 1,2,....k
k
k
d dD then D for k
dx dx .Then the equation (1) can be expressed as
10 1
10 1
b ... ( )
. ., (b ... ) ( )
n nn
n nn
D y b D y b y R x
i e D b D b y R x
i.e., f(D) y = R(i) where f(D) = 1
0 1b ...n nnD b D b is called a differential operator. To
solve such equations we first study the properties of the differential operator.
1.13.4 Properties of differential operator:
1. f(D)eax
= eax
f(a)
Proof: Let f(D) = 1
0 1b ...n nnD b D b .
Since Dke
ax = a
ke
ax, for k = 1,2,3…n. we have,
ax 10 1
10 1
10 1
f(D)e = b ...
...
( ... ) ( )
n ax n ax axn
n ax n ax axn
n n ax axn
D e b D e b e
b a e b a e b e
b a b a b e f a e
2. f(D)eax
y = eax
f(D+a)y
Proof: Let f(D) = 1
0 1b ...n nnD b D b
We have D eax
y = y Deax
+ eax
Dy = y eax
a + eax
Dy = eax
(D+a)y,
D2(e
ax y) = D(D(e
ax y)) = D(e
ax (D+A)y) = e
ax (D+a)(D+a)y= e
ax (D+a)
2y
Similarly we can show that Dk(e
ax y) = e
ax (D+a)
ky, k = 1,2,...n.
Substituting in the formula we get the required result.
3. 0, 0,1,2,... 1
( )!,
k ax j
ax
j kD a e x
e k j k
Proof: We know that 0, 0,1,2,.... 1
!,
k j j kD x
k j k
then by property(2), the result follows.
1.13.5 The solution of linear homogeneous differential equation with constant coefficients
Consider the linear homogeneous differential equation with constant coefficients
1
0 1 11... 0
n n
n nn n
d y d y dyb b b b y
dx dx dx
where are all constants.
21
This equation can be rewritten as f(D)y = 0 where f(D) = 10 1b ...n n
nD b D b .
If y = axe , then by property (1), f(D)y = f(D)eax
= eax
f(a) = 0 f(a) = 0.
This equation is called the auxiliary or characteristic equation associated with the given
differential equation. For a nth
order differential equation, the auxiliary equation has n roots say,
1 2, ,...... .na a a Then 1 21 2, ,..........., na xa x a x
ny e y e y e are all solutions of the given differential
equation.
Case1: If the roots of the auxiliary equations are all distinct then
1 21 2, ,..........., na xa x a x
ny e y e y e are linearly independent solutions of the given equation.
Hence the general solution is given by 1 21 2 ... na xa x a x
ny C e C e C e .
Case2: If some roots are equal, say 1 2 ...... .ka a a a
Then the solutions 1 2 ... axky y y e . Then the solution is given by 1 2 ... na xa x a x
y e e e
does not contain n arbitrary constants and hence cannot be the general solution. Since first k
roots of the auxiliary equations are equal the given differential equation can be rewritten as
( )( ) 0.kg D D a y Then by property (3), we observe that , 0,1,......,. 1ax jjy e x j k are all
solutions of the given equation. Hence the general solution is
12 11 2 3 1( ...... ) ......k na x a xk ax
k k ny C C x C x C x e C e C e .
Case3. If some roots of the auxiliary equation are real. Since for equation with real coefficients
the roots exists in conjugate pairs, let 1 2a iband a ib be two roots. Then
1 2( ) ( )1 2(cos sin ) (cos sin )
x xa ib x ax a ib x axy e e e bx i bx and y e e e bx i bx are the two
distinct solutions. Therefore the corresponding solution is 1 21 2
x xy C e C e , which can be
expressed as 1 1 2( cos sin )axy e A bx A bx where 1 2A and A are arbitrary constants.
Example 1.13.4: Solve (2D2 + 5D – 12)y = 0
Solution: Auxiliary equation is 2m2 + 5m – 12 = 0.
That is, (2m – 3) (m + 4) = 0 and it has the roots m1 = 3/2 and m2 = - 4 which are real and
distinct. Therefore the general solution is y = C1 e3x/2
+ C2e-4x
Example 1.13.5 Solve 2
2
dx
yd + 4 y = 0
22
Solution: Auxiliary equation is m2 + 4 = 0 or m
2= - 4
Therefore m = 2i = 0 2i
Therefore y = C1 cos 2x + C2 sin 2x
Example 1.13.6 Solve 3
3
dx
xd + 4
2
2
dx
yd + 4
dx
dy = 0.
Solution: Auxiliary equation is m3 + 4m
2 + 4m = 0.
That is, m(m + 2)2 = 0
Therefore m1 = 0, m2 = -2, m3 = -2 are its roots.
Here we find that two roots are equal. Hence the general solution is given by
y = C1 e0x
+ (C2 + C3x) e-2x
Or y = C1 + (C2 + C3i) e-2x
Example 1.13.7 Solve4
4
dx
xd + 8
2
2
dx
yd + 16y = 0
Solution: Here the auxiliary equation is m4 + 8m
2 + 16 = 0, which is simply (m
2 + 4)
2 = 0.
Therefore we have m2 + 4 = 0 repeated. It gives m
2 = -4
Therefore m = 2i.
Thus the roots are m = 2 i, 2i (imaginary, repeated).
Hence the general solution is y = e0x
[(C1 + C2x) cos 2x + (C3 + C4x) sin 2x]
That is, y = (C1 + C2x) cos 2x + (C3 + C4x) sin 2x.
Example 1.13.8 Solve 4
4
dx
yd - 2
3
3
dx
yd + 2
2
2
dx
yd - 2
dx
dy + y = 0
Solution: Hence the A.E. is m4 – 2m
3 + 2m
2 – 2m + 1 = 0.
The roots are m = 1, 1, i
Therefore the general solution is y = (C1 + C2x) ex + e
0x (C3 cos x + C4 sin x)
That is, y = (C1 + C2x)ex + (C3 cos x + C4 sin x).
1.13.6 The solution of linear non-homogeneous differential equation with constant
coefficients
We know that the general solution of nonhomogeneous linear differential equation
23
is of the form c py y y , where cy is the general solution of the corresponding homogenous
equation , called the complementary function and
py is a particular solution of the given equation , which does not contain any arbitrary constants.
The complementary function can be determined using the method described above. To determine
the particular solution py , we use the following methods.
1.13.6.1 Inverse differential operator method
If f(D)y = (x) then we define the inverse differential operator denoted by 1
( )f D as
)(
1
Df [ (x)] = y , where ‘D’ is the differential operator
dx
d.
Thus f(D) is also a differential operator and )(
1
Df can be treated as its inverse.
Example 1.13.6.1 1
( ) ( ) .y ydx DR x y R x ydxD
2
2
1 1 1 1 11 1. , 1 1
2
11 , 1,2,3,.......
!
k
k
xdx x x xdx
D D D DD
xSimilarly wecan showthat k
kD
1.13.6.2. Properties of the inverse differential operator:
2. 1
( ) 0.( ) ( )
axax e
e if f af D f a
Proof: We know that f(D)eax
= eax
f(a), If f(a) 0, then dividing by f(a) we get
1( ) ( )
( ) ( )
1
( ) ( )
axax
axax
ef D e f a
f a f a
ee
f D f a
3. 1 1
( ) ( )
ax axe y e yf D a f D
Proof: From the property f(D)eax
y = eax
f(D+a)y the result follows.
4. 1 ax axy e e y dx
D a
Proof: The result follows directly from the result (2) and example (1) .
24
To determine the particular solution of a linear non-homogenous differential equation, we use
inverse differential operators. If f(D)y = (x) ,then yp = )(
1
Df [ (x)]
Case(i): If (x) = eax
, then yp = ( )
axe
f a if f(a) 0.
If f(a) = 0, then f(D) can expressed as f(D) = (D-a)k (D), where (a) 0.for k = 1,2,3,….
k k k
k k
1 1 1 1 1 Then
f(D) (D) (a)(D-a) (D) (D-a) (D-a)
1 1 11
(a) (a) (a) !(D-a) D
axax ax ax
ax ax kax
ee e e
e e xe
k
Case(ii): If (x)= cosax or sinax then, since cosax = Re(iaxe ) and sinax = Im(
iaxe ) this case
reduces to the case(i) and hence can be solved.
Case(iii) If (x)= xm, for some positive integer m, then
1
f(D) can be expanded as a series in
positive powers of x and hence 1
f(D)
mx can be determined.
Working Rule:
1. If (x) = eax
, then
2
2
1( ) 0.
( ) ( )
1 1( ) 0 ( ) 0,
( ) ( ) ( )
1( ) 0.
( )
1( ) 0 ...
( )
axax
axax ax
ax
ax
ee if f a
f D f a
eIf f a then e x e x if f a
f D f D f a
x e if f af D
x e if f a and so onf a
2. If (x) = cos ax or sin ax and if the differential operator can be written as f(D2) then
)(
12Df
[(x)] = 2
1
( )f a (x) provided f(-a
2) 0.
If f(-a2) = 0 and 2
2 2
1 1f (-a ) 0, then ( (x)) = x. ( (x))
f(D ) f (-a )
25
Example 1.13.8 Solve 2
2
dx
yd - 6
dx
dy + 10 y = cos 2x + e
-3x
Solution: Auxiliary equation is m2 – 6m + 10 = 0.
Therefore m = 1.2
40366 =
6 4
2
=
6 2
2
i = 3 i
Or m = i, where = 3 and = 1.
Therefore C.F. = e3x
(C1 cos x + C2 sin x)
P.I. - 106
12 DD
(cos 2x) + 106
12 DD
(e-3x
)
Using P.I rule 2 and rule 1 respectively we get,
P.I. = 1062
12 D
(cos 2x) + 10)3(6)3(
12
e-3x
= D66
1
(cos 2x) +
10189
1
.e
-3x
= 6
121
1
D
D
(cos 2x) +
37
1e
-3x, multiplying numerator and denominator by 1 + D in the first
expression.
That is, P.I. = 6
1
)2(1
12
D(cos 2x) +
37
1e
-3x
30
1{1 . cos 2x + D(cos 2x)} +
37
1 e
-3x.
That is, P.I = 30
1(cos 2x – 2 sin 2x) +
37
1 e
-3x, since D =
dx
d
General solution is y = C.F. + P.I.
Therefore y = e3x
(C1 cos x + C2 sin x) + 30
1 (cos 2x – 2 sin 2x) +
37
1 e
-3x
Example 1.13.9 Solve (D4 + 18 D
2 + 81) y = cos
3 x.
Solution: Auxiliary equation is m4 + 18 m
2 + 81 = 0
That is (m2 + 9)
2 = 0. Therefore m = 3i, 3i.
Thus C.F. = e0.x
{(C1 + C2 x) cos 3x + (C3 + C4x) sin 3x}
C.F. = (C1 + C2x) cos 3x + (C3 + C4x) sin 3x.
26
P.I = 8118
124 DD
(cos3 x)
= 8118
124 DD
xx 3cos
4
1cos
4
3.
( Since Cos3 A =
4
3 cos A +
4
1 cos 3A).
P.I = 4
3
8118
124 DD
(cos x) + 4
14 2
1
18 81D D (cos 3x)
= 4
3
81)1(18)1(
1222
(cos x) + 4
1
8118
124 DD
(cos 3x)
But f(D2) = D
4 + 18D
2 + 81 and a = 3 in the second expression.
We observe that f(-a2) = f(-9) = 81 – 162 + 81 = 0
Also f 1
(D) = 4D3 + 36 D + 0 = 4 D(D
2 + 9)
Therefore f 1
(-9) = 4 D(-9 + 9) = 0, also
Hence 8118
124 DD
(cos 3x) = x2
)(
1211 Df
(cos 3x)
But f 1
(D) = 4D3 + 36 D, Therefore f
11(D) = 12 D
2 + 36
Hence 8118
124 DD
(cos 3x) = x2,
)3(
1211 f
cos 3x
That is, x2
36)9(12
1
cos 3x = -
72
1 x
2 cos 3x
Thus P.I. = 4
3.
81181
1
cos x +
4
1
xx 3cos
72
1 2
Therefore P.I. = 256
3.cos x -
288
1.x
2 cos 3x.
That is, y = (C1 + C2x) cos 3x + (C3 + C4x) sin 3x + 256
3 cos x -
288
1x
2 cos 3x.
Example 1.13.10 Solve (D2 – 6D + 9)y = x
2 + x + 1
Solution: Auxiliary equation m2 – 6m + 9 = 0.
That is (m – 3)2 = 0. Therefore m = 3, 3
Thus C.F. = (C1 + C2 x)e3x
.
27
P.I = 96
12 DD
(x2 + x + 1)
That is, =
2
9
1
3
219
1
DD
(x2 + x + 1)
= 9
1.
1
2
9
1
3
21
DD (x
2 + x + 1)
(Using the binomial expansion (1 – a)-1
= 1 + a + a2 + …), we have
P.I = 9
1
...
9
1
3
2
9
1
3
21
2
22 DDDD (x2 + x + 1)
= 9
1{1 +
3
2 D -
9
1D
2 +
9
4D
2 + higher powers} (x
2 + x + 1)
= 9
1(1 +
3
2D +
3
1D
2)(x
2 + x+ 1), neglecting D
3 and higher powers since we have only 2
nd degree
polynomial x2 + x + 1.
Therefore P.I. = 9
1{1(x
2 + x + 1) +
3
2
dx
d(x
2 + x + 1) +
3
12
2
dx
d(x
2 + x + 1)}
= 9
1{(x
2 + x + 1) +
3
2(2x + 1) +
3
1(2)}
= 9
1x
2 +
27
7x +
27
7 =
27
1 (3x
2 + 7x + 7)
Thus y = C.F. + P. I. = (C1 + C2x) e3x
+ 27
1(3x
2 + 7x + 7).
Example 1.13.11 Solve 3
3
dx
yd + 3
2
2
dx
yd = 1 + x + e
-3x
Solution: Auxiliary equation is m3 + 3m
2 = 0
Thus C.F. = (C1 + C2x) e0x
+ C3 e-3x
or C1 + C2 x + C3 e-3x
P.I. = 23 3
1
DD (1 + x + e
-3x)
=
313
1
2 DD
(1 + x + e-3x
)
28
= 23
1
D
1
31
D(1 + x) + x .
DD 63
12
e-3x
= 23
1
D
2 3
1 ...3 9 27
D D D
(1 + x) + x.
DD 63
12
e-3x
= 2
1 1 1
3 9 27 81
D
D D
(1 + x) + x.
)3(6)3(3
12
e-3x
= 3
1.
2
1
D(1 + x) -
9
1.
D
1(1 + x) +
27
1(1 + x) -
81
1D(1 + x) +
9
xe
-2x
But D
1(1 + x) = x +
2
2
x,
2
1
D(1 + x) =
2
2x +
6
3x and D(1 + x) = 1.
Therefore P.I. = 3
1
62
32 xx -
9
1
9
2xx +
27
1 (1 + x) -
81
1 (1) +
9
xe
-3x
= 81
2 -
27
2x +
25
162x
2 +
1
18 x
3 +
9
xe
-3x
The term 81
2 -
27
2x may be neglected in view of the arbitrary C1 + C2x appearing in C.F.
General Solution is y = C.F. + P.I. = C1 + C2x + C3 e-3x
+ 25
162x
2 +
1
18 x
3 +
9
xe
-3x
Example 1.13.12 Solve 3
3
dx
yd + 2
2
2
dx
yd +
dx
dy = e
2x + x
3
Solution: Auxiliary equation is m3 + 2m
2 + m = 0
m(m2 + 2m + 1) = 0 or m (m + 1)
2 = 0
Therefore m = 0, -1, -1
Thus C.F. = C1 e0x
+ (C2 + C3 x) e-x
or C1 + (C2 + C3 x) e-x
P.I. = DDD 23 2
1(e
2x) +
DDD 23 2
1(x
3)
Now, DDD 23 2
1(e
2x) =
22.22
123
.e2x
= 18
2xe
Consider DDD 23 2
1(x
3)
Here f(D) = D3 + 2 D
2 + D and f(0) = 0
29
Therefore we write, DDD 23 2
1(x
3) =
)12(
12 DDD
(x3)
= D
1.
2)1(
1
D(x
3) =
D
1(1 + D)
-2 (x
3)
(Using (1 + a)-2
= 1 – 2 a + 3a2 - + …), we have
DDD 23 2
1(x
3) =
D
1(1 – 2D + 3D
2 – 4D
3 + 5 D
4) (x
3), neglecting higher powers of D.
= (D
1 - 2 + 3D – 4D
2 + 5 D
3) (x
3)
Now, D
1 stands for integration, and D, D
2 etc. for successive derivatives.
Therefore DDD 23 2
1(x
3) =
4
4x - 2 .x
3 + 3(3x
2) – 4 (6x) + 5(6).
= 4
4x - 2x
3 + 9x
2 – 24x + 30
Thus P.I. = 18
1 e
2x +
4
4x - 2x
3 + 9x
2 – 24x + 30
The constant term 30 may be neglected in view of the arbitrary constant C1 in the C.F.
Hence, the general solution y = C1 + (C2 e-x
– C3) e-x
+ 18
1e
2x +
4
4x - 2x
3 + 9x
2 – 24x.
Exercise: 1.13.13
1. Solve 4 2
2
dx
yd + 16
dx
dy - 9y = 4 e
x/2 + 3 sin (x/4)
2. Solve (D2 + 1)y = e
x + x
4 + sin x.
Answers :
1. Auxiliary equation is 4m2 + 16 m – 9 = 0. The roots are m =
2
1 and -
2
9 are the roots.
The C.F. = C1 ex/2
+ C2 e-9x/2
P.I. = 9164
12 DD
[4ex/2
+ 3 sin (x/4)]
= 4 9164
12 DD
(ex/2
) + 3. 9164
12 DD
[sin (x/4)].
30
In the first expression f(D) = 4 D2 + 16 D – 9.
Therefore f
2
1 = 4
2
2
1
+ 16
2
1 - 9 = 1 + 8- 9 = 0.
Therefore 9164
12 DD
(ex/2
) = x.168
1
D(e
x/2).
In the second expression, replace D2 by -
24
1.
Therefore P.I. = 4x 168
1
D(e
x/2) + 3.
9164
14
1
2
D
[sin (x/4)]
P.I = 4x.
162
18
1
.e
x/2 + 3.
9164
1
1
D
[sin (x/4)]
= 5
1xe
x/2 + 3.
2
2
4
37256
4
3716
D
D
[sin (x/4)]
Thus P.I = 5
1xe
x/2 +
2
2 4
37
4
1256
4
37163
D
[sin (x/4)]
= 5
1 xe
x/2 -
1625
12(64 D + 37) sin (x/4)
General solution is
y = C.F + P.I. = C1 ex/2
+ C2 e-9x/2
+ 5
1xe
x/2 -
1625
12{16 cos (x/4) + 37 sin (x/4)}
2. The roots of the auxiliary equation is m = i.
Therefore C.F. = e0x
(C1 cos x + C2 sin x) that is, = C1 cos x + C2 sin x.
P.I = 1
12 D
(ex + x
4 + sin x) =
2
1e
x + (x
4 – 12x
2 +24) + x .
2
1(-cos x )
General solution is y = C1 cos x + C2 sin x + 2
1 e
x + x
4 –12 x
2 + 24 -
2
x cos x.
1.14 Variation of Parameters
31
Consider the second order linear differential equation
( ) ( ) ( ). (1)y p x y q x y R x
Suppose that 1 1 2 2cy c y c y is a solution, where
1 2 and y y are linearly independent on an
interval .a x b Let us see what happens if we replace both of the constants 1 2 c and c with
functions of x.
We consider 1 2 (2)y Ay By and try to determine ( ) ( )A x and B x so that
1 2Ay By is a solution of the equation (1).
Then 1 2 1 2. (3)y Ay By A y B y
Rather than involved with the derivatives of ( ) ( )A x and B x of higher order than the first, we
now choose some particular function for the expression 1 2A y B y .
For simplicity we choose 1 2 0. (4)A y B y
It then follows from (3) that 1 2 1 2 . (5)y Ay By A y B y
Since y was to be a solution of (1) we substitute from (2), (3), and (5) into equation (1) to obtain
1 1 1 2 2 2 1 2( ) ( ) ( ).A y py qy B y py qy A y B y R x
But 1 2 and y y are solutions of the homogeneous equation, so that finally
1 2 ( ). (6)A y B y R x
Equations (4) and (6) now give us two equations that we wish to solve for ( ) ( ).A x and B x
This solution exists providing the determinant 1 2
1 2
( )y y
W xy y
does not vanish. But this
determinant is precisely the Wronskian of the functions1 2 and y y , which are presumed to be
linearly independent on the interval .a x b Therefore, the Wronskian does not vanish on that
interval and we can find .A and B By integration we get
2 1( ) ( )( ) ( )
( ) ( )
y R x y R xA x dx and B x dx
W x W x
Example 1.14.1 Solve the equation 2( 1) sec tan .D y x x
Immediately we find that 1 2cos sin .cy c x c x
32
Let us seek a particular solution by variation of parameters. Put cos sin .py A x B x
Then
1 2
1 2
cos sin( ) 1.
sin cos
y y x xW x
x xy y
22 ( ) sin sec tan( ) tan tan .
( ) 1
y R x x x xA x dx dx xdx x x
W x
(Constant of integration has been disregarded because we are seeking only a particular solution.)
And
1 ( ) cos sec tan( ) tan ln | sec | .
( ) 1
y R x x x xB x dx dx xdx x
W x
Therefore, the complete solution is
1 2
1 3
( ) cos sin cos ( tan ) sin ln | sec |
cos sin cos sin ln | sec |,
y x c x c x x x x x x
c x c x x x x x
where the term ( sin )x in py has been absorbed in the complementary function term
3 sinc x ,
since 3c is an arbitrary constant.
Example 1.14.2 Solve 2 1
( 3 2) .1 x
D D ye
Here 2
1 2 ,x x
cy c e c e so we put
2 .x x
py Ae Be
Then
21 2 3
2
1 2
( ) .2
x x
x
x x
y y e eW x e
e ey y
2
2
-x 3x
( )( ) ln(1 ).
( ) (1+e )e 1
x xx
x
y R x e eA x dx dx dx e
W x e
And
2
1 ( )( ) ln(1 ).
( ) (1 ) 1
x xx x x
x x
y R x e eB x dx dx e dx e e
W x e e
33
Therefore, the complete solution is 2 2
3 2( ) ( )ln(1 ).x x x x xy x c e c e e e e
Exercises 1.14.3
Solve using variation of parameters:
1. 2( 1) csc cot .D y x x
2. 2 4( 1) sec .D y x
3. 2 2( 1) tan .D y x
4. 2 2( 1) sec csc .D y x x
5. 2 2 2( 2 1) ( 1) .x xD D y e e
6. 2( 3 2) cos( ).xD D y e
7. 2 2 1/ 2( 1) 2(1 ) .xD y e
8. 2 2( 1) sin .x xD y e e
1.14 Cauchy’s homogeneous linear equation:
This equation is of the form
1
1
1 11... ................ 1
n nn n
n nn n
d y d y dyx k x k x k y X
dx dx dx
, where X is a function of x , and ik ,
1,2..., ,i n are constants. Equations of this type can be reduced to linear differential equations
with constant coefficients by letting tx e . Thus logt x .
If ,d
Ddt
then 1
. .dy dy dt dy
dx dt dx dt x ; i.e.,
dyx Dy
dx
2 2
2 2 2
1 1.
d y d dy d y dy
dx dx x dt x dt dt
; i.e.,
22
21
d yx D D y
dx .
Similarly, 3
3
31 2
d yx D D D y
dx , and so on.
After making these substitutions in equation (1), we get a linear equation with constant
coefficients which can be solved as before.
Example 1.15.1 Solve 2
2 2
24 6
d y dyx x y x
dx dx
Solution: Put tx e . Then logt x . Let ,d
Ddt
then dy
x Dydx
, 2
2
21
d yx D D y
dx .
The given equation becomes 21 4 6 tD D D y e ; i.e., 2 25 6 tD D y e
The auxiliary equation is 2 5 6 0m m ; i.e., 2 3 0m m .
34
The roots are 2,3m . The complementary function is 2 3
1 2
t t
cy c e c e
The particular integral is 2
2
1
5 6
t
py eD D
21
2 5
tt eD
2
2 2 5
tet
2tte
The complete solution is 2 3 2
1 2 logc py y y c x c x x x
Example 1.15.2 Solve 2
2 3
22 12 log
d y dyx x y x x
dx dx
Solution: Put tx e . Then logt x . Let ,d
Ddt
then dy
x Dydx
, 2
2
21
d yx D D y
dx .
The given equation becomes 2 312 tD D y te
The roots of the auxiliary equation are 3, 4m
The complementary function is 3 4
1 2
t t
cy c e c e
The particular integral is 3
2
1
12
t
py teD D
3 2 1 1
7 2 7 49
te tt
The complete solution is
233 4
1 2
log 1 1log
7 2 7 49c p
xxy y y c x c x x
Example 1.15.3 Solve 3 2
3 2
3 23 log
d y d y dyx x x y x x
dx dx dx
Solution: Put tx e . Then logt x .
Let ,d
Ddt
then dy
x Dydx
, 2
2
21
d yx D D y
dx ,
33
31 2
d yx D D D y
dx
The given equation becomes 3 1 tD y e t
The roots of the auxiliary equation are 1 3
1,2
im
;
The complementary function is 21 2 3
3 3cos sin
2 2
tt
cy c e e c t c t
The particular integral is 3
1
1
t
py e tD
2
tet
The complete solution is
1
1 2 3
3 3cos log sin log log
2 2 2c p
xy y y c x x c x c x x
Exercise 1.15.4
Solve
1. 2
2 4
22 4
d y dyx x y x
dx dx
35
2. 2
2 2
2 1d yx y x
dx x x
3. 2
2
2sin log log
d y dyx x y x x
dx dx
Answers
1. 4
1 4
1 2 log5
c p
xy y y c x c x x
2. 2 2
1 2
1 log 1
3
xy c x c x
x x
3. 2
1 2
1 1cos log sin log log cos log log log sin log
4 4c py y y c x c x x x x x
1.16 Legendre’s linear equation:
This equation is of the form
1
1
1 11... ................ 1
n nn n
n nn n
d y d y dyax b k ax b k ax b k y X
dx dx dx
,
where X is a function of x , and ik , 1,2..., ,i n are constants. Equations of this type can be
reduced to linear differential equations with constant coefficients by letting tax b e . Thus
logt ax b .
If ,d
Ddt
then 1
. . .dy dy dt dy
adx dt dx dt ax b
; i.e., dy
ax b aDydx
2 2
221
d y aD D y
dx ax b
; i.e.,
22 2
21
d yax b a D D y
dx .
Similarly, 3
3 3
31 2
d yax b a D D D y
dx , and so on.
After making these substitutions in equation (1), we get a linear equation with constant
coefficients which can be solved as before.
Example 1.16.1 Solve 2
2
22 3 2 3 12 6
d y dyx x y x
dx dx
Solution: Put 2 3 tx e . Then log 2 3t x .
Let ,d
Ddt
then 2 3 2dy
x Dydx
, 2
2 2
22 3 2 1
d yx D D y
dx .
The given equation becomes 3
4 1 2 12 62
teD D D y
;
i.e., 24 6 12 3 9tD D y e
The roots of the auxiliary equation are 3 57
4m
;
The complementary function is 3 57 3 57
4 41 2
t t
cy c e c e
36
The particular integral is 2
13 9
4 6 12
t
py eD D
3 3
14 4
te
The complete solution is 3 57 3 57
4 41 2
3 32 3 2 3 2 3
14 4c py y y c x c x x
Example 1.16.2 Solve 3 2
3 2
3 21 2 1 4 1 4 4log 1
d y d y dyx x x y x
dx dx dx
Solution: Put 1 tx e . Then log 1t x . Let ,d
Ddt
then 1dy
x Dydx
,
2
2
21 1
d yx D D y
dx ,
33
31 1 2
d yx D D D y
dx
The given equation becomes 3 2 4 4 4D D D y t
The roots of the auxiliary equation are 1,2, 2m ;
The complementary function is 2 2
1 2 3
t t t
cy c e c e c e
The particular integral is 3 2
14
4 4py t
D D D
12
1 ( ...) ( ) 14
DD t t
The complete solution is 2 2
1 2 31 1 1 log 1 1c py y y c x c x c x x
Example 1.16.3 Solve 2
2
21 1 sin 2log 1
d y dyx x y x
dx dx
Solution: Put 1 tx e . Then log 1t x .
Let ,d
Ddt
then 1dy
x Dydx
, 2
2
21 1
d yx D D y
dx .
The given equation becomes 2 1 sin 2D y t
The roots of the auxiliary equation are m i ;
The complementary function is 1 2cos sincy c t c t
The particular integral is 2
1sin 2
1py t
D
1sin 2
3t
The complete solution is
1 2
1cos log 1 sin log 1 sin 2log 1
3c py y y c x c x x
Exercise 1.16.4
Solve
1. 2
2 2
23 2 5 3 2 3 1
d y dyx x y x x
dx dx
2. 2
2
21 1 2sin log 1
d y dyx x y x
dx dx
3. 2
2 2
22 1 2 1 2 8 2 3
d y dyx x y x x
dx dx
Answers
37
1. 1 1 2
31 2
1 7 13 2 3 2 3 2 3 2
405 27 108y c x c x x x
2. 1 2coslog 1 sin log 1 log 1 cos log 1y c x c x x x
3. 1 2
21 2
1 12 1 2 1 2 1 2 1 log 2 1 2
5 2y c x c x x x x
1.17 System of linear differential equations with constant coefficients
Quite often we come across a system of linear differential equations with constant coefficients in
which there are two or more dependent variables and a single independent variable exists. Such
a system of equation can be solved by eliminating all but one of the dependent variables , and
solving the resulting equation. Then using the given equations, the other dependent variables
can be expressed in terms of the dependent variable which is obtained earlier and can be
determined.
Example 1.17.1 Solve the simultaneous equations:
5 2 ; 2 0dx dy
x y t x ydt dt
being given 0x y when 0.t
Solution: Taking ,d
Ddt
the given equations become
( 5) 2 ( )
2 ( 1) 0 ( )
D x y t i
x D y ii
Eliminate x as if D were an ordinary algebraic multiplier. Multiplying (i) by 2 and operating on
(ii) by (D+5) and then subtracting, we get
2( 6 9) 2 .D D y t
Its complementary function is 3
1 2( ) ( ) t
cy t c c t e and a particular integral is
2
1 2 4( ) ( 2 ) .
( 3) 9 27p
ty t t
D
Thus ( ) .c py t y y
When t=0, 1
40 .
27y c
Substituting the value of y in (ii), we obtain
3
2 2
4 1 1( ) .
27 2 9 27
t tx t c c t e
When t = 0, 2
20 .
9x c
Hence the desired solutions are
3 31 1 2 2( ) (1 6 ) (1 3 ); ( ) (2 3 ) (2 3 ).
27 27 27 27
t tx t t e t y t t e t
38
Example 1.17.2 Solve the simultaneous equations
2 sin 0; 2 cos 0dx dy
y t x tdt dt
given that x = 0 and y = 1 when t = 0.
Solution: Taking ,d
Ddt
the given equations become
2 sin ( )
2 cos ( )
Dx y t i
x Dy t ii
Eliminating x by multiplying (i) by 2 and operating on (ii) by D and then adding, we get
2( 4) 3sin .D y t
Its complementary function is 1 2( ) cos2 sin2cy t c t c t and a particular integral is
2
1( ) 3 sin sin .
4py t t t
D
Thus ( ) .c py t y y
When t=0, 11 1.y c
Substituting the value of y in (ii), we obtain
2( ) sin2 cos2 cos .x t t c t t
When t = 0, 20 1.x c
Hence the desired solutions are
( ) cos2 sin2 cos ; ( ) cos2 sin2 sin .x t t t t y t t t t
Example 3:
Solve the simultaneous equations
2 2cos 7sin ; 2 4cos 3sin .dx dy dx dy
y t t x t tdt dt dt dt
Solution: Taking ,d
Ddt
the given equations become
( 2) 2cos 7sin ( )
( 2) 4cos 3sin ( )
Dx D y t t i
D x Dy t t ii
Eliminating y by operating on (i) by D and operating on (ii) by (D-2) and then adding, we get 2( 2) 9cos .D x t
Its complementary function is 2 2
1 2( ) t t
cx t c e c e and a particular integral is
( ) 3cos .px t t
Thus ( ) .c px t x x
Substituting the value of x in (ii), we obtain
39
2 2
1 2 3( ) ( 2 1) ( 2 1) 2sin .t tx t c e c e t c
Where 1 2 3, and c c c are arbitrary constants.
Exercise:
Solve
i. sin 0; cos 0dx dy
y t x tdt dt
given that x = 2 and y = 0 when t = 0.
ii. ( 1) 2 1;(2 1) 2D x Dy t D x Dy t .
iii. ( 1) (2 1) ;( 1) ( 1) 1tD x D y e D x D y .
iv. 2( 1) 4( 1) 4 ;( 1) ( 9) 0xD y D v e D y D v given y = 5, dy/dx = 0, v=1/2
at x = 0. ( Hint: Eliminate v(x) first and solve for y(x))
2. The small oscillations of a certain system with two degrees of freedom are given by the
equations 2 2 2( 3) 2 0;( 3) ( 5) 0D x y D x D y where
dD
dt . If x = y =
0, Dx=3, Dy = 2 when t = 0, find x and y when t = ½.
3. A mechanical system with two degrees of freedom satisfies the equations 2 2
2 22 3 4;2 3 0.
d x dy d y dx
dt dt dt dt Obtain expressions for x and y in terms of t, given
x, y, dx/dt, dy/dt all vanish at t = 0.
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