Concepts of Higher Order Linear Differential Equations - (3.1)
1. Higher Order Linear Differential Equations:
An nth-order linear differential equation:
dny
d n"1 y
dy
a n Ÿx n a n"1 Ÿx n"1 . . . a 1 Ÿx a 0 Ÿx y gŸx dx
dx
dx
dny
d n"1 y
dy
Let LŸy a n Ÿx n a n"1 Ÿx n"1 . . . a 1 Ÿx a 0 Ÿx y. L is a linear operator meaning for any
dx
dx
dx
constant ) and *
LŸ)y *z )LŸy *LŸz .
An initial-value problem for an nth-order linear differential equation: solve
dny
d n"1 y
dy
a 0 Ÿx y gŸx a n Ÿx n a n"1 Ÿx n"1 . . . a 1 Ÿx dx
dx
dx
subject to yŸa y 0 , y U Ÿa y 1 , . . . , y Ÿn"1 Ÿa y n"1 .
A boundary-value problem for a 2nd-order linear differential equation: solve
d2y
dy
a 2 Ÿx 2 a 1 Ÿx a 0 Ÿx y gŸx dx
dx
subject to yŸa ),
yŸb *.
Boundary-value problems with other boundary conditions:
yŸa ), y U Ÿb *
y U Ÿa ), yŸb *
y U Ÿa ), y U Ÿb *
c 1 yŸa c 2 y U Ÿa ), c 3 yŸb c 4 y U Ÿb *.
Homogeneous and nonhomogeneous differential equations:
Homogeneous: gŸx 0, and LŸy 0
Nonhomogeneous: gŸx p 0, and LŸy gŸx 2. Existence of a Unique Solution of An Initial-Value Problem:
Let a 0 Ÿx , a 1 Ÿx , . . . , a n Ÿx and gŸx be continuous on an interval I, and let a n Ÿx p 0 for every x in I. If
x a is a point in I, then the initial-value problem has a unique solution in I.
Example Ÿ1 Show that yŸx 3 C 1 x C 2 x 2 is the general solution of the differential equation
x 2 y UU " 2xy U 2y 6
Ÿ2 Find an interval I on which the initial-value problem
x 2 y UU " 2xy U 2y 6, yŸ1 3, y U Ÿ1 1
has a unique solution; and Ÿ3 solve the initial value problem.
Ÿ1 Verify that y satisfies the differential equation:
y U C 1 2C 2 x,
y UU 2C 2
x 2 y UU " 2xy U 2y x 2 Ÿ2C 2 " 2xŸC 1 2C 2 x 2Ÿ3 C 1 x C 2 x 2 x 2 Ÿ2C 2 " 4C 2 2C 2 xŸ"2C 1 2C 1 6 6
Ÿ2 Here a 0 Ÿx 2, a 1 Ÿx "2x, a 2 Ÿx x 2 , and gŸx 6.
Since a 2 Ÿx x 2 0, x 0, consider I 0, . . Then a 0 Ÿx , a 1 Ÿx , a 2 Ÿx , and gŸx 6 are
continuous on I and a 2 Ÿx p 0 for every x in I. The initial value of x 1 is also in I. So, the
1
initial-value problem has a unique solution on I.
Ÿ3 Solve constants C 1 and C 2 so that y satisfies the initial conditions.
yŸ1 3 C 1 C 2 3
U
y Ÿ1 C 1 2C 2 1
C1
C2
®
C1 C2 0
®
C 1 2C 2 1
1 1
1 2
"1
0
1
the solution of the initial value problem:
1 1
C1
1 2
C2
0
1
"1
1
y 3 " x x2
Example Ÿ1 Show that y C 1 cos 4x C 2 sin 4x is the general solution of the differential equation
y UU 16y 0. Ÿ2 Solve the boundary-value problems
Ÿi yŸ0 0, y =
2
UU
y 16y 0, Ÿii yŸ0 0, y =
8
Ÿiii yŸ0 0, y =
2
Ÿ1 Verify that y satisfies the differntial equation:
y U "4C 1 sin 4x 4C 2 cos 4x,
0;
0;
1.
y UU "16C 1 cos 4x " 16C 2 sin 4x
y UU 16y "16C 1 cos 4x " 16C 2 sin 4x 16ŸC 1 cos 4x C 2 sin 4x 0
Ÿ2 Solve the boundary value problems:
i. yŸ0 0, y = 0
2
yŸ0 C 1 0, y = C 2 sinŸ2= 0, C 2 can be any real number
2
So, the BVP has infinitely many solutions:
y C 2 sin 4x.
=
ii. yŸ0 0, y
0
8
yŸ0 0, C 1 0, y = C 2 sin = C 2 0
8
2
So, the BVP has a unique solution: y 0.
iii. yŸ0 0, y = 1
2
yŸ0 0, C 1 0, y = C 2 sinŸ2= 0 p 1
2
So, the BVP has no solution.
3. Homogeneous Linear Differential Equations:
a. The general solution of a homogeneous nth-order linear differential equation:
Let y 1 , . . . , y k be solutions of the homogeneous nth-order differential equation LŸy 0 on an interval
I. Then the linear combination:
y C 1 y 1 C 2 y 2 . . . C n y n
U
where C i s are arbitrary constants, is also a solution on I.
It is easy to see. Since LŸy i 0, for i 1, . . . , n,
LŸy LŸC 1 y 1 C 2 y 2 . . . C n y n C 1 LŸy 1 . . . C n LŸy n 0.
b. Linear independence of solutions:
2
A set of functions: f 1 Ÿx , . . . , f n Ÿx is said to be linearly independent on an interval I if
C 1 f 1 Ÿx C 2 f 2 Ÿx . . . C n f n Ÿx 0 implies C 1 C 2 . . . C n 0
for all x in I. If a set is not linearly independent on I, then it is said to be linearly dependent.
Example The set of functions 1, x, x 2 , . . . , x n is linearly independent on "., . .
2
1
-4
-2
0
2 x
-4
4
4
20
2
10
0
-2
-1
2 x
-4
4
0
-2
-2
-10
-4
-20
2 x
4
-2
yx
y1
y x2
None of them can be a linear combination of others. So, 1, x, x 2 , . . . , x n is linearly independent on
"., . .
Example Determine if the set of functions: 1, sin 2 x, cos 2x, is linearly independent on "., . .
Since sin 2 x 12 Ÿ1 " cos 2x 12 " 12 cos 2x, a linear combination of 1 and cos 2x, the set of
1, sin 2 x, cos 2x is linearly dependent on "., . .
2
1
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
-4
-2
0
2 x
4
-4
-2
-1
-2
y1
2 x
4
0
-0.2
-2
-0.4
-0.4
-0.6
-0.6
-0.8
-0.8
-1
-1
0, 1 . Since x 0, |x| x,
ii. Consider x in I 2 "1, 1 . Then x C 1 x C 2 |x| 2 x
x, |x|
x
if x u 0
"x
if x 0
x, x
is linearly independent on
is linearly dependent on
and
C 1 x C 2 x xŸC 1 C 2 if x u 0
C 1 x " C 2 x xŸC 1 " C 2 if x 0
C1 C2 0
if x u 0
C1 " C2 0
if x 0
4
y cos 2x
0, 1 , I 2 Ÿ"1, 1 . Determine if x, |x|
i. Consider x in I 1 I1.
C 1 x C 2 |x| 0 ®
-4
y sin 2 x
Example Let x, |x| and I 1 i. I 1 and ii. I 2 .
3
0
-0.2
®
1
1
C1
1 "1
C2
.
0
0
C1
C2
x, |x|
1
1
"1
0
1 "1
0
0
0
is linearly independent on I 2 .
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
-1 -0.8 -0.6 -0.4 -0.2
-0.2
0.2 0.4 x 0.6 0.8
0
-1 -0.8 -0.6 -0.4 -0.2
-0.2
1
-0.4
-0.4
-0.6
-0.6
-0.8
-0.8
-1
-1
0.2 0.4 x0.6 0.8
1
yx
y |x|
c. Linear independence of solutions of a linear differential equation:
Definition: Wronskian:
Let functions f 1 Ÿx , . . . , f n Ÿx have at least n " 1 derivatives. The determinant
WŸf 1 , . . . , f n det
f1
f2
...
fn
f U1
f U2
...
f Un
:
:
:
:
f Ÿn"1 . . . f Ÿn"1 f Ÿn"1 n
1
2
is call the Wronskian of the functions: f 1 Ÿx , . . . , f n Ÿx .
Criterion for linearly independent solutions:
Let y 1 , y 2 , . . . , y n be n solutions of the homogeneous linear nth-order differential equation on an
interval I. Then the set of solutions is linearly independent on I if and only if
WŸy 1 , . . . , y n p 0 for every x in I.
It can be seen as follows. Let C 1 y 1 . . . C n y n 0. Solve for C Ui s. Since
Ÿk C 1 y Ÿk 1 . . . C n y n 0, for k 0, 1, . . . , n " 1,
y1
y2
...
yn
C1
y U1
y U2
...
y Un
C2
:
:
:
:
:
y Ÿn"1 . . . y Ÿn"1 y Ÿn"1 n
1
2
0
0
:
0
Cn
C1
WŸy 1 , . . . , y n p 0 for every x in I ·
C2
:
Cn
0
0
:
0
Example Determine if the set of solutions of a differential equation is linearly independent I, specify I.
Ÿii e )x , e *x , ) p *
Ÿi 1, x, x ln x
4
y U1 0
y1 1
i.
y U2 1
,
y2 x
y UU1 0
,
y U3 ln x x 1x
y 3 x ln x
ln x 1
y UU2 0
y UU3 1x
1 x x ln x
0 1 ln x 1
0 0 1x
W 1, x, x ln x is defined on either 0, . or "., 0 . Let I Hence, 1, x, x ln x is linearly independent on I.
W 1, x, x ln x
ii.
y 1 e )x
y 2 e *x
,
det
1x
0, . . W 1, x, x ln x
p 0.
y U1 )e )x
y U2 *e *x
W e )x , e *x
det
Since * p ) and e Ÿ)* x p 0,
e )x
)e
)x
e )x , e *x
e *x
*e
*x
*e Ÿ)* x " )e Ÿ)* x e Ÿ)* x Ÿ* " ) is linearly independent on "., . .
d. Fundamental set of solutions:
A set y 1 , . . . , y n of an nth-order linearly independent solutions of a homogeneous differential
equation: LŸy 0 on I is said to be a fundamental set of solutions.
Existence of a Fundamental Set of Solutions:
There exists a fundamental set of solutions for a homogeneous linear nth-order differential equation
LŸy 0.
The General Solution of a Linear nth-order Homogeneous Differential Equation: LŸy 0
Let y 1 , . . . , y n be a fundamental set of solutions of LŸy 0 for x in I. Then the general solution of the
equation LŸy 0 for x in I is
y C 1 y 1 . . . C n y n
U
where C i s are arbitrary constants.
Example i. Show that y 1 e x and y 2 e "x/2 are solutions of the differential equation
2y UU " y U " y 0
ii. Find the general solution of the equation.
i. Verify y 1 and y 2 are solutions of the differential equation.
y 1 e x , y U1 e x , y UU1 e x , 2y UU1 " y U1 " y 1 2e x " e x " e x 0
y 2 e "x/2 , y U2 " 1 e "x/2 , y UU2 1 e "x/2
2
4
2 1 e "x/2 " " 1 e "x/2 " e "x/2 1 e "x/2 1 e "x/2 " e "x/2 0
2
2
2
4
x
"x/2
is linearly independent on "., . . y 1 e x
ii. From the previous example, we know e , e
"x/2
and y 2 e
form a fundamental set of solutions. Hence, the general solution is
y C 1 e x C 2 e "x/2 .
4. Nonhomogeneous Linear Differential Equations:
5
a. Solution of nonhomogeneous equation:
Consider LŸy gŸx . A function y p satisfy the equation LŸy gŸx is called a particular solution
of the equation: LŸy gŸx .
General solution of a nonhomogeneous equation: LŸy gŸx Let y 1 , . . . , y n be a fundamental set of solutions of the homogeneous equation: LŸy 0. Then the
general solution of LŸy gŸx for x in I is
y C 1 y 1 . . . C n y n y p
where C Ui s are arbitrary constants.
b. If y p and y q are particular solutions of LŸy gŸx and LŸy hŸx , respectively, then the general
solution of the equation LŸy gŸx hŸx is
y C 1 y 1 . . . C n y n y p y q
Example Ÿ1 Show that y p 1
5
cos x "
3
sin x
5
UU
U
and y q "3x 2 6x " 18 are solutions of the equations
2y " y " y 2 sin x,
2y UU " y U " y 3x 2
respectively. Ÿ2 Find a particular solution of the equation 2y UU " y U " y 2 sin x 3x 2 .
Ÿ1 Verify that y p satisfies the differential equation: 2y UU " y U " y 2 sin x
y Up " 1 sin x " 3 cos x, y UUp " 1 cos x 3 sin x
5
5
5
5
2y UUp " y Up " y p " 2 cos x 6 sin x 1 sin x 3 cos x " 1 cos x 3 sin x
5
5
5
5
5
5
cos x " 2 3 " 1 sin x 6 1 3 2 sin x
5
5
5
5
5
5
2
UU
U
satisfies
the
differential
equation:
2y
Verify that y q
" y " y 3x
U
UU
y q "6x 6, y q "6
2y UU " y U " y 2Ÿ"6 " Ÿ"6x 6 " Ÿ"3x 2 6x " 18 x 2 Ÿ3 xŸ6 " 6 Ÿ"12 " 6 18 3x 2
Ÿ2 Let y r y p y q 15 cos x " 35 sin x "3x 2 6x " 18 . y r is a particular solution of the differential
equation: 2y UU " y U " y 2 sin x 3x 2 .
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