Analytical Solutions to Ordinary Differential Equations 5

August 28, 2013

Higher-Order LDE

Linear Differential Equations (LDE)

The dependent variable and its derivatives appear to

the first degree only, and the coefficients are either

constants or functions of the independent variable only.

Examples: 3

3+

2

2+

+ = tanh

3

3+ 2

2

2+ 7

+ 3 = 0

Higher-Order LDE

Linear Differential Equations (LDE)

Examples: 3

3+

2

2+

+ = tanh

3

3+ 2

2

2+ 7

+ 3 = 0

Variable Coefficient Constant Coefficient

Higher-Order LDE

Linear Differential Equations (LDE)

Examples: 3

3+

2

2+

+ = tanh

3

3+ 2

2

2+ 7

+ 3 = 0

Non-Homogeneous DE Homogeneous DE

Outline

1. Solution to Homogeneous LDE with Constant Coefficients

2. Solution to Non-homogeneous LDE with Constant Coefficients A. Method of Undetermined Coefficients

B. Method of Variation of Parameters

3. Solution to Special Higher-Order LDE with

Variable Coefficients

LDE with Constant Coefficients

General Form:

)(... 012

2

21

1

1 xfyadx

dya

dx

yda

dx

yda

dx

yda

n

n

nn

n

n

To ease in writing the equation, let:

=

=

+ 1

1 + + 22 + 1 + 0 = ()

= () is a polynomial in D

LDE with Constant Coefficients

= () is a polynomial in D

Properties of the D operator:

1. + = +

2. + + = + +

3. =

4. =

5. + = +

6. = +

7. = k = constant

Outline

1. Solution to Homogeneous LDE with Constant Coefficients

2. Solution to Non-homogeneous LDE with Constant Coefficients A. Method of Undetermined Coefficients

B. Method of Variation of Parameters

3. Solution to Special Higher-Order LDE with

Variable Coefficients

I. Solution to Homo LDE w/ CC

General Form: = 0

Example: 2 + 1 = 0

2

2+ = 0 OR

2 + 1 = 0 1. Characteristic equation:

2. Solve for the roots

Case 1: Real & Distinct Case 2: Complex & Distinct Case 3: Real & Repeated Case 4: Complex & Repeated

I. Solution to Homo LDE w/ CC

General Form: = 0

Characteristic Equation: = 0

Case 1: Real and Distinct Roots

Solution: = 11 + 2

2 + +

Example: Solve completely 3 42 + + 6 = 0

Solution: = 1 + 2

2 + 33

I. Solution to Homo LDE w/ CC

General Form: = 0

Characteristic Equation: = 0

Case 2: Complex and Distinct Roots

Solution: = 11 + 2

2 + +

Example: Solve completely 2 + 1 = 0

Recall: = cos sin

= 1 cos + 2 sin

I. Solution to Homo LDE w/ CC

General Form: = 0

Characteristic Equation: = 0

Case 3: Real and Repeated Roots

Solution: = 1 + 2 + + 1

Example: Solve completely 2 + 6 + 9 = 0

Solution: = 1 + 2 3

I. Solution to Homo LDE w/ CC

General Form: = 0

Characteristic Equation: = 0

Case 4: Complex and Repeated Roots

Solution: =

1 + 2 + + 1 cos

+ +1 + + 21 sin

Example: Solve completely 4 + 22 + 1 = 0

Solution: = 1 + 2 cos + 3 + 4 sin

I. Solution to Homo LDE w/ CC

Example: Solve completely:

= 2 3, 2 3, 4, 4, 4, 2, 2

2 4 + 13 2 4 3 + 2 2 + 4 = 0

= 2 1 + 2 cos 3 + 3 + 4 sin 3 +

5 + 6 + 72 4 + 8

2 + 9 cos 2 + 10 sin 2

Roots of the characteristic equation:

Final Answer:

I. Solution to Homo LDE w/ CC

Example: Solve completely:

2 4 + 13 2 4 3 + 2 2 + 4 = 0

Yes! = can be a solution to homogeneous LDEs with either constant or variable coefficients. But

This happens only if

, , , =

Outline

1. Solution to Homogeneous LDE with Constant Coefficients

2. Solution to Non-homogeneous LDE with Constant Coefficients A. Method of Undetermined Coefficients

B. Method of Variation of Parameters

3. Solution to Special Higher-Order LDE with

Variable Coefficients

II. Solution to Non-homo LDE w/ CC

The complete solution to an n-th order LDE:

= (, 1, 2, , )

If LDE is non-homo, it always has two solutions:

= Complementary Function

= Particular Integral

By the Principle of Superposition, the general solution is:

= +

II. Solution to Non-homo LDE w/ CC

1. Complementary Function

a. contains all arbitrary constants

b. satisfies the Homo LDE

2. Particular Integral

a. must not contain arbitrary constants

b. satisfies given LDE

= 0

= ()

By the Principle of Superposition, the general solution is:

= +

Solution to Homo LDE with CC

Solution by MUC or MVP

II. A. MUC (Undetermined Coefficients)

*We can solve certain forms of f(x) only.

= ()

f(x) yP

(constant) (constant)

0 + 1 + 22 + +

0 + 1 + 22 + +

cos or sin cos + sin

cos or sin

0 + 1 + 22 + +

cos + 0 + 1 + 2

2 + + sin

II. A. MUC (Undetermined Coefficients)

Steps:

1. Check whether f(x) is one of those in the table

2. Solve for yc using the previous lesson.

3. Solve for yp, note that yp = yp1 + yp2 + + ypk

4. Compare yc & yp, whenever any term in yp duplicates

any term in yc, that yp term must be multiplied by the

lowest possible integer power of x sufficient to

eliminate duplication.

5. Combine like terms

6. Substitute assumed form of yp into DE and evaluate

coefficients of identical terms.

II. A. MUC (Undetermined Coefficients)

Example: Solve completely using MUC.

3 + 62 + 11 + 6 = + cos

Solution:

= 1 + 2

2 + 33

= + cos + sin

Roots: = 1, 2, 3 Write all yP:

II. A. MUC (Undetermined Coefficients)

Solve for A, B, and C:

Roots: = 1, 2, 3 Write all yP:

= 1 + 2

2 + 33

= + cos + sin

= sin + cos

3 = + sin cos

2 = cos sin

= + cos

Compute A, B, C such that

=1

24 = 0 =

1

10

= 1 + 2

2 + 33 +

1

24 +

1

10sin

Outline

1. Solution to Homogeneous LDE with Constant Coefficients

2. Solution to Non-homogeneous LDE with Constant Coefficients A. Method of Undetermined Coefficients

B. Method of Variation of Parameters

3. Solution to Special Higher-Order LDE with

Variable Coefficients

II. B. MVP (Variation of Parameters)

Steps:

1. Solve for yc

2. Replace cs with us to get yp

3. Differentiate yp as many times as the order of

the DE. Set the sum of all terms containing

derivatives of us equal to zero except in last

differentiation.

4. Solve for u1 , u2, , un

5. Obtain u1, u2, un by integration to get final yp

II. B. MVP (Variation of Parameters)

Example: Solve completely using MVP.

2 + 1 = csc cot

Solution:

= 1 cos + 2 sin = 1 cos + 2 sin

= 1 cos 1 sin + 2

sin + 2 cos

2 = 1 sin 1 cos + 2

cos 2 sin

II. B. MVP (Variation of Parameters)

Example: Solve completely using MVP.

2 + 1 = csc cot

Solution:

= 1 cos + 2 sin cos ln sin cos sin

cos sin sin cos

1

2 =

0csc cot

In order to solve for u1 and u2:

Outline

1. Solution to Homogeneous LDE with Constant Coefficients

2. Solution to Non-homogeneous LDE with Constant Coefficients A. Method of Undetermined Coefficients

B. Method of Variation of Parameters

3. Solution to Special Higher-Order LDE with

Variable Coefficients

This is the Last Slide

+ 4 = 3 csc

6 = 103

12 + 48 64 = 12 328 + 24

HOMEWORK. Solve these LDE individually. This is due next meeting. Use MVP for the first item, and MUC for the last 2. Submit on yellow paper.