Nonhomogeneous Linear Systems of Differential...

Nonhomogeneous Linear Systems of

Differential Equations:

the method of variation of parameters

Xu-Yan Chen

◮ Nonhomogeneous Linear Systems of DifferentialEquations:

(∗)nh

d~x

dt= A(t)~x +~f(t)


(∗)nh

d~x

dt= A(t)~x +~f(t)

◮ No general method of solving this class of equations.


(∗)nh

d~x

dt= A(t)~x +~f(t)


◮ Solution structure: The general solutions of the nonhomogsystem (∗)nh are of the form:

~x(t) = ~xp(t) + ~xc(t),

where ~xp(t) is a particular solution of of nonhomog system (∗)nh,and ~xc(t) are general solutions of the homog system

(∗)h

d~xc

dt= A(t)~xc.


(∗)nh

d~x

dt= A(t)~x +~f(t)



~x(t) = ~xp(t) + ~xc(t),


(∗)h

d~xc

dt= A(t)~xc.

◮ What is this note about?


(∗)nh

d~x

dt= A(t)~x +~f(t)



~x(t) = ~xp(t) + ~xc(t),


(∗)h

d~xc

dt= A(t)~xc.

◮ What is this note about?Variation of Parameters.


(∗)nh

d~x

dt= A(t)~x +~f(t)



~x(t) = ~xp(t) + ~xc(t),


(∗)h

d~xc

dt= A(t)~xc.


If the solutions ~xc(t) ofthe homogeneous system(∗)h have been providedor prepared,


(∗)nh

d~x

dt= A(t)~x +~f(t)



~x(t) = ~xp(t) + ~xc(t),


(∗)h

d~xc

dt= A(t)~xc.


If the solutions ~xc(t) ofthe homogeneous system(∗)h have been providedor prepared,

⇒we can solve the nonhomogsystem (∗)nh.

Fundamental Matrices of Homogeneous Linear Systems:


◮ Suppose that the solutions of the homog sys ~xc′ = A(t)~xc are:

~xc(t) = C1~x1(t) + C2~x2(t),

where ~x1(t) and ~x2(t) are two linearly independent solutions.



~xc(t) = C1~x1(t) + C2~x2(t),


◮ The matrix M(t) =[

~x1(t) ~x2(t)]

is called a fundamental

matrix.



~xc(t) = C1~x1(t) + C2~x2(t),



~x1(t) ~x2(t)]


matrix.

◮ The general solutions of ~xc′ = A(t)~xc can be expressed as:

~xc(t) = M(t)~C.



~xc(t) = C1~x1(t) + C2~x2(t),



~x1(t) ~x2(t)]


matrix.


~xc(t) = M(t)~C.

Since

~xc(t) = C1~x1(t) + C2~x2(t) =

[

~x1(t) ~x2(t)

] [

C1

C2

]

= M(t)~C



~xc(t) = C1~x1(t) + C2~x2(t),



~x1(t) ~x2(t)]


matrix.


~xc(t) = M(t)~C.



~xc(t) = C1~x1(t) + C2~x2(t),



~x1(t) ~x2(t)]


matrix.


~xc(t) = M(t)~C.

◮ The solution of the I.V.P. ~xc′ = A(t)~xc, ~xc(t0) = ~a:

~xc(t) = M(t)M(t0)−1 ~a.



~xc(t) = C1~x1(t) + C2~x2(t),



~x1(t) ~x2(t)]


matrix.


~xc(t) = M(t)~C.


~xc(t) = M(t)M(t0)−1 ~a.

Since

~xc(t0) = M(t0)~C = ~a ⇒ ~C = M(t0)−1 ~a

⇒ ~xc(t) = M(t)M(t0)−1 ~a.



~xc(t) = C1~x1(t) + C2~x2(t),



~x1(t) ~x2(t)]


matrix.


~xc(t) = M(t)~C.


~xc(t) = M(t)M(t0)−1 ~a.



~xc(t) = C1~x1(t) + C2~x2(t),



~x1(t) ~x2(t)]


matrix.


~xc(t) = M(t)~C.


~xc(t) = M(t)M(t0)−1 ~a.

◮ M ′(t) = A(t)M(t).



~xc(t) = C1~x1(t) + C2~x2(t),



~x1(t) ~x2(t)]


matrix.


~xc(t) = M(t)~C.


~xc(t) = M(t)M(t0)−1 ~a.

◮ M ′(t) = A(t)M(t).

◮ M(t) is an invertible matrix for any t.



~xc(t) = C1~x1(t) + C2~x2(t),



~x1(t) ~x2(t)]


matrix.


~xc(t) = M(t)~C.


~xc(t) = M(t)M(t0)−1 ~a.

◮ M ′(t) = A(t)M(t).

◮ M(t) is an invertible matrix for any t.

◮ The fundamental matrices are not unique.

Nonhomogeneous Linear Systems of Differential Equations:

(∗)nh ~x′ = A(t)~x +~f(t).

The Method of Variation of Parameters:


(∗)nh ~x′ = A(t)~x +~f(t).


◮ Suppose that the homogeneous system ~x′c = A(t)~xc is solved,

with

{

a fundamental matrix M(t),

the complementary solutions ~xc(t) = M(t)~C.


(∗)nh ~x′ = A(t)~x +~f(t).



with

{



◮ Set ~x(t) = M(t)~u(t).


(∗)nh ~x′ = A(t)~x +~f(t).



with

{



◮ Set ~x(t) = M(t)~u(t).

◮ Substitute this in (∗)nh. It simplifies to ~u ′(t) = M(t)−1~f (t).


(∗)nh ~x′ = A(t)~x +~f(t).



with

{



◮ Set ~x(t) = M(t)~u(t).


Proof:`

M(t)~u(t)´′

= A(t)M(t)~u(t) +~f(t)

⇒ M ′(t)~u(t) + M(t)~u ′(t) = A(t)M(t)~u(t) +~f(t).

Since M ′(t) = A(t)M(t), we obtain M(t)~u ′(t) = ~f(t).

Take the inverse: ~u ′(t) = M(t)−1 ~f(t).


(∗)nh ~x′ = A(t)~x +~f(t).



with

{



◮ Set ~x(t) = M(t)~u(t).


Proof:`

M(t)~u(t)´′

= A(t)M(t)~u(t) +~f(t)




◮ Integrate to get ~u(t).


(∗)nh ~x′ = A(t)~x +~f(t).



with

{



◮ Set ~x(t) = M(t)~u(t).


Proof:`

M(t)~u(t)´′

= A(t)M(t)~u(t) +~f(t)




◮ Integrate to get ~u(t).

◮ Finally, multiply to get ~x(t) = M(t)~u(t).

Example 1: Solve ~x ′ =

[

1 14 1

]

~x +

[

−1 + 21t + e3t tan t

2 + 30t + 2e3t tan t

]


[

1 14 1

]

~x +

[

−1 + 21t + e3t tan t

2 + 30t + 2e3t tan t

]

◮ Eigenvalues & eigenvectors of

[

1 14 1

]

(details skipped here)

⇒ Complementary solutions: ~xc(t) = C1e−t

[

−12

]

+ C2e3t

[

12

]

⇒ A fundamental matrix: M(t) =

[

−e−t e3t

2e−t 2e3t

]


[

1 14 1

]

~x +

[

−1 + 21t + e3t tan t

2 + 30t + 2e3t tan t

]


[

1 14 1

]



[

−12

]

+ C2e3t

[

12

]


[

−e−t e3t

2e−t 2e3t

]

◮ Set ~x(t) = M(t)~u(t)


[

1 14 1

]

~x +

[

−1 + 21t + e3t tan t

2 + 30t + 2e3t tan t

]


[

1 14 1

]



[

−12

]

+ C2e3t

[

12

]


[

−e−t e3t

2e−t 2e3t

]

◮ Set ~x(t) = M(t)~u(t)

◮ The original system is simplified to ~u ′(t) = M(t)−1~f(t):

~u ′(t) =

[

−e−t e3t

2e−t 2e3t

]−1 [

−1 + 21t + e3t tan t

2 + 30t + 2e3t tan t

]

=1

−4e2t

[

2e3t −e3t

−2e−t −e−t

] [

−1 + 21t + e3t tan t

2 + 30t + 2e3t tan t

]

=

[

(1 − 3t)et

18te−3t + tan t

]

.

Example 1 (continued): ~x ′ =

[

1 14 1

]

~x +

[

−1 + 21t + e3t tan t

2 + 30t + 2e3t tan t

]

◮ Integrate ~u ′(t) =

[

(1 − 3t)et

18te−3t + tan t

]

:

R

(1 − 3t)etdt = (1 − 3t)et −R

(−3)etdt = (1 − 3t)et + 3et + C,R

18te−3tdt = −6te−3t −R

(−6)e−3tdt = −6te−3t − 2e−3t + C,R

tan t dt = − ln | cos t| + C.

⇒ ~u(t) =

[

(4 − 3t)et + C1

(−2 − 6t)e−3t − ln | cos t| + C2

]

.

Example 1 (continued): ~x ′ =

[

1 14 1

]

~x +

[

−1 + 21t + e3t tan t

2 + 30t + 2e3t tan t

]


[

(1 − 3t)et

18te−3t + tan t

]

:

R

(1 − 3t)etdt = (1 − 3t)et −R

(−3)etdt = (1 − 3t)et + 3et + C,R

18te−3tdt = −6te−3t −R

(−6)e−3tdt = −6te−3t − 2e−3t + C,R

tan t dt = − ln | cos t| + C.

⇒ ~u(t) =

[

(4 − 3t)et + C1

(−2 − 6t)e−3t − ln | cos t| + C2

]

.

◮ Finally, the solutions ~x(t) are obtained from ~x(t) = M(t)~u(t):

~x(t) =

[

−e−t e3t

2e−t 2e3t

] [

(4 − 3t)et + C1

(−2 − 6t)e−3t − ln | cos t| + C2

]

=

−6 − 3t − e3t ln | cos t|

4 − 18t − 2e3t ln | cos t|

+ C1e−t

[

−12

]

+ C2e3t

[

12

]

.

Example 2: Solve 2y′′ − y′ − y =3e

3

2t

1 + et.


3

2t

1 + et.

◮ Set ~x =

[

x1

x2

]

=

[

y

y′

]

:

2y′′ − y′ − y = 3e3

2t

1+et ⇔ ~x ′ =

[

0 112

12

]

~x +

[

03e

3

2t

2(1+et)

]


3

2t

1 + et.

◮ Set ~x =

[

x1

x2

]

=

[

y

y′

]

:

2y′′ − y′ − y = 3e3

2t

1+et ⇔ ~x ′ =

[

0 112

12

]

~x +

[

03e

3

2t

2(1+et)

]

◮ Characteristic roots ⇒ yc(t) = C1e− 1

2t + C2e

t


3

2t

1 + et.

◮ Set ~x =

[

x1

x2

]

=

[

y

y′

]

:

2y′′ − y′ − y = 3e3

2t

1+et ⇔ ~x ′ =

[

0 112

12

]

~x +

[

03e

3

2t

2(1+et)

]


2t + C2e

t

◮ yc(t) ⇒ complem. solutions of the system & a fund. matrix:

~xc(t) = C1

[

e−1

2t

− 12e−

1

2t

]

+ C2

[

et

et

]

, M(t) =

[

e−1

2t et

− 12e−

1

2t et

]

.


3

2t

1 + et.

◮ Set ~x =

[

x1

x2

]

=

[

y

y′

]

:

2y′′ − y′ − y = 3e3

2t

1+et ⇔ ~x ′ =

[

0 112

12

]

~x +

[

03e

3

2t

2(1+et)

]


2t + C2e

t

◮ yc(t) ⇒ complem. solutions of the system & a fund. matrix:

~xc(t) = C1

[

e−1

2t

− 12e−

1

2t

]

+ C2

[

et

et

]

, M(t) =

[

e−1

2t et

− 12e−

1

2t et

]

.

◮ Set ~x(t) = M(t)~u(t). The system simplifies to:

~u ′(t) = M(t)−1~f(t) =

[

e−1

2t et

− 12e−

1

2t et

]−1[

03e

3

2t

2(1+et)

]

=1

32e

1

2t

[

et −et

12e−

1

2t e−

1

2t

]

[

03e

3

2t

2(1+et)

]

=

− e2t

1+et

e1

2t

1+et

.

Example 2 (continued): 2y′′ − y′ − y = 3e3

2t

1+et


− e2t

1+et

e1

2t

1+et

:

R

e2t

1+et dt =R

s1+s

ds =R

(1 − 1

1+s)ds = s − ln |1 + s| + C

(substituted s = et),R

e1

2t

1+etdt =

R

2

1+s2ds = 2arctan s + C (substituted s = et/2).

⇒ ~u(t) =

[

−et + ln(1 + et) + C1

2 arctan(e1

2t) + C2

]

.

Example 2 (continued): 2y′′ − y′ − y = 3e3

2t

1+et


− e2t

1+et

e1

2t

1+et

:

R

e2t

1+et dt =R

s1+s

ds =R

(1 − 1

1+s)ds = s − ln |1 + s| + C

(substituted s = et),R

e1

2t

1+etdt =

R

2

1+s2ds = 2arctan s + C (substituted s = et/2).

⇒ ~u(t) =

[

−et + ln(1 + et) + C1

2 arctan(e1

2t) + C2

]

.

◮ Finally, the solutions ~x(t) are obtained from ~x(t) = M(t)~u(t):

~x(t) =

[

e−1

2t et

− 12e−

1

2t et

] [

−et + ln(1 + et) + C1

2 arctan(e1

2t) + C2

]

= · · · ,

y(t) = x1(t)

= −e1

2t + e−

1

2t ln(1 + et) + 2et arctan(e

1

2t) + C1e

− 1

2t + C2e

t.

Example 3: Solve (1 + t)y′′ + (−1 − 2t)y′ + ty =et

1 + t,

given that y0(t) = et satisfies (1 + t)y′′0 + (−1 − 2t)y′

0 + ty0 = 0.


1 + t,


0 + ty0 = 0.

• Reduction of Order: y0(t) ⇒ yc(t).• Variation of Parameters: yc(t) ⇒ Solutions of the nonhomog eq.


1 + t,


0 + ty0 = 0.


• Reduction of Order: Solve (∗)h (1+ t)y′′c +(−1− 2t)y′

c + tyc = 0


1 + t,


0 + ty0 = 0.



c + tyc = 0

◮ Set yc(t) = y0(t)z(t) = etz(t) and substitute y(t) = etz(t) in (∗)h:

(1 + t)(etz)′′ + (−1 − 2t)(etz)′ + tetz = 0,

(1 + t)(etz′′ + 2etz′ + etz) + (−1 − 2t)(etz′ + etz) + tetz = 0,


1 + t,


0 + ty0 = 0.



c + tyc = 0


(1 + t)(etz)′′ + (−1 − 2t)(etz)′ + tetz = 0,


(1 + t)etz′′ + etz′ = 0,


1 + t,


0 + ty0 = 0.



c + tyc = 0


(1 + t)(etz)′′ + (−1 − 2t)(etz)′ + tetz = 0,


(1 + t)etz′′ + etz′ = 0,

(∗)z (1 + t)z′′ + z′ = 0.


1 + t,


0 + ty0 = 0.



c + tyc = 0


(∗)z (1 + t)z′′ + z′ = 0.


1 + t,


0 + ty0 = 0.



c + tyc = 0


(∗)z (1 + t)z′′ + z′ = 0.

◮ Set w(t) = z′(t). The eq (∗)z becomes a 1st order linear eq:

(∗)w (1 + t)w′ + w = 0.


1 + t,


0 + ty0 = 0.



c + tyc = 0


(∗)z (1 + t)z′′ + z′ = 0.


(∗)w (1 + t)w′ + w = 0.

◮ Solve (∗)w ⇒ w(t) =C1

1 + t.


1 + t,


0 + ty0 = 0.



c + tyc = 0


(∗)z (1 + t)z′′ + z′ = 0.


(∗)w (1 + t)w′ + w = 0.

◮ Solve (∗)w ⇒ w(t) =C1

1 + t.

◮ z′ = w ⇒ z(t) =∫

w(t)dt =∫ C1

1 + tdt = C1 ln |1 + t| + C2


1 + t,


0 + ty0 = 0.



c + tyc = 0


(∗)z (1 + t)z′′ + z′ = 0.


(∗)w (1 + t)w′ + w = 0.

◮ Solve (∗)w ⇒ w(t) =C1

1 + t.

◮ z′ = w ⇒ z(t) =∫

w(t)dt =∫ C1

1 + tdt = C1 ln |1 + t| + C2

◮ yc(t) = y0(t)z(t) = etz(t) = et [C1 ln |1 + t| + C2].

Or, equivalently, yc(t) = C1et ln |1 + t| + C2e

t.

Example 3 (continued): (1 + t)y′′ + (−1 − 2t)y′ + ty =et

1 + t.

• Reduction of Order: y0(t) = et ⇒ yc(t) = C1et ln |1 + t| + C2e

t.

• Variation of Parameters: yc(t) ⇒ Solve the nonhomog eq.


1 + t.


t.


◮ yc(t) gives a fundamental matrix:

M(t) =

[

y1 y2

y′1 y′

2

]

=

[

et ln |1 + t| et

et ln |1 + t| + et

1+tet

]


1 + t.


t.



M(t) =

[

y1 y2

y′1 y′

2

]

=

[

et ln |1 + t| et

et ln |1 + t| + et

1+tet

]

◮ ~x =

[

x1

x2

]

=

[

y

y′

]

satisfies ~x ′ = A(t)~x +

[

0et

(1+t)2

]

.


1 + t.


t.



M(t) =

[

y1 y2

y′1 y′

2

]

=

[

et ln |1 + t| et

et ln |1 + t| + et

1+tet

]

◮ ~x =

[

x1

x2

]

=

[

y

y′

]

satisfies ~x ′ = A(t)~x +

[

0et

(1+t)2

]

.

◮ Set ~x(t) = M(t)~u(t). Then ~u(t) satisfies ~u ′(t) = M(t)−1~f(t).

~u ′(t) =

[

et ln |1 + t| et

et ln |1 + t| + et

1+tet

]−1[

0et

(1+t)2

]

=1

− e2t

1+t

[

et −et

−et ln |1 + t| − et

1+tet ln |1 + t|

]

[

0et

(1+t)2

]

=

[

11+t

− ln |1+t|1+t

]

.


1 + t.

Variation of Parameters (continued)


[

11+t

− ln |1+t|1+t

]

:

R

1

1+tdt = ln |1 + t| + C,

R ln |1+t|1+t

dt =R

sds = 1

2s2 + C (substituted s = ln |1 + t|).

⇒ ~u(t) =

[

ln |1 + t| + C1

− 12 (ln |1 + t|)2 + C2

]

.


1 + t.

Variation of Parameters (continued)


[

11+t

− ln |1+t|1+t

]

:

R

1

1+tdt = ln |1 + t| + C,

R ln |1+t|1+t

dt =R

sds = 1

2s2 + C (substituted s = ln |1 + t|).

⇒ ~u(t) =

[

ln |1 + t| + C1

− 12 (ln |1 + t|)2 + C2

]

.

◮ Finally, we can get ~x(t) = M(t)~u(t) and y(t) = x1(t):

y(t) = u1(t)et ln |1 + t| + u2(t)e

t

= ln |1 + t| · et ln |1 + t| − 12 (ln |1 + t|)

2· et

+C1et ln |1 + t| + C2e

t

= 12et (ln |1 + t|)

2+ C1e

t ln |1 + t| + C2et.

Nonhomogeneous Linear Systems of Differential...

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