High Order Differential Equations National Chiao Tung University Chun-Jen Tsai 10/2/2019

High Order Differential Equations

National Chiao Tung University Chun-Jen Tsai 10/2/2019

Initial-Value Problems

 For a linear n^th-order differential equation, an initial- value problem (IVP) is:

Solve:

Subject to:

) ( )

( )

( ...

) ( )

( _{1 1 0}

1

1 a x y g x

dx x dy dx a

y x d

dx a y x d

a _n

n n n

n

n  _ ^_    

1 0

) 1 ( 1

0 0

0) , ( ) ,..., ( )

(x  y y x  y y ^n x  y_n y

Existence of a Unique Solution

 Theorem: Let a_n(x), a_n-1(x), …, a₁(x), a₀(x) and g(x) be continuous on an interval I, and let a_n(x)  0 for every x in this interval. If x = x₀ is any point in this interval, then a solution y(x) of the IVP exists on the interval and is unique.

Examples:

 The trivial solution y = 0 is the unique solution of the IVP 3y⁽³⁾ + 5y" – y' + 7y = 0, y(1) = y'(1) = y"(1) = 0 on any interval containing x = 1.

 The solution y = 3e^2x + e^–2x – 3x is the unique solution of the IVP y"– 4y = 12x, y(0) = 4, y'(0) = 1 on any interval containing x = 0.

 The solution family y = cx² + x + 3 are solutions of the IVP x²y"– 2xy' + 2y = 6, y(0) = 3, y'(0) = 1

 a₂(x) = x² = 0 at 0.

Boundary-Value Problem

 Solving a linear DE with y or its derivatives specified at different points. For example,

Solve

Subject to

) ( )

( )

( )

( _{2 1 0}

2

2 a x y g x

dx x dy dx a

y x d

a   

1 0, ( )

)

(a y y b y

y  

y

(a, y₀)

(b, y₁) Solutions of the DE

Solutions of a BVP

 A BVP can have many, one, or no solutions.

Example: x = c₁ cos 4t + c₂ sin 4t is a solution family of x" + 16x = 0. What are the solutions of the BVPs with (1) x(0) = 0, x(



(2) x(0) = 0, x(



(3) x(0) = 0, x(



x

t (0, 0)

1

-1

C₂= 1

C₂= 1/2 C₂= 1/4 C₂= 0

C₂= –1/2

(/2, 0)

Homogeneous Equations

 For a linear n^th-order differential equation

if g(x) = 0, it is called a homogeneous differential equation, otherwise, it is non-homogeneous.

Note that: the solution of a non-homogeneous

differential equation is based on the solution to its associated homogeneous differential equation.

), ( )

( )

( ...

) ( )

( _{1 1 0}

1

1 a x y g x

dx x dy dx a

y x d

dx a y x d

a _n

n n n

n

n  _ ^_    

Differential Operators

 The symbol D, defined by Dy = dy/dx, is called a

differential operator. D transforms a function into another function.

Example: D(cos 4x) = –4sin 4x, D(5x³ – 6x²) = 15x² – 12x

 Polynomial expressions involving D, such as D + 3 and D² + 3D – 4 are also differential operators.

Linear Operator

 An n^th-order differential operator is defined as:

L = a_n(x)Dⁿ＋a_n-1(x)D^n–1＋…＋a₁(x)D＋a₀(x) L is a linear operator, that is,

L{









“D” Representation of DEs

 Any differential equations can be expressed in terms of the D notation.

For example, y" + 5y' + 6y = 5x – 3 can be written as D²y＋5Dy＋6y = 5x－3 or (D²＋5D＋6)y = 5x－3

 A linear n^th-order differential equation can be write compactly as L(y) = g(x).

Superposition Principle

 Theorem: Let y₁, y₂,…,y_k be solutions of a homogeneous linear n^th-order DE on an interval I. Then the linear

combination

y = c₁y₁(x) + c₂y₂(x) +… + c_ky_k(x),

where c_i, i = 1, 2, …, k are arbitrary constants, is also a solution on this interval.

 Can be proved by using linear operator property.

Linear Dependency

 A set of functions f₁(x), f₂(x), …, f_n(x) is said to be linearly dependent on an interval I if there exist constants c₁, c₂,…, c_n, not all zero, such that

c₁ f₁(x) + c₂ f₂(x) +…+ c_n f_n(x) = 0.

Otherwise, it’s said to be linearly independent.

 Example: Are cos²x, sin²x, sec²x, tan²x linearly dependent on the interval (–





Wronskian

 We are interested in linearly independent solutions of a linear differential equations  How to verify?

 Suppose each of the functions f₁(x), f₂(x),…, f_n(x) has at least n–1 derivatives. The determinant

is called the Wronskian of the functions.

  

Criterion for Linear Independency

 Theorem: Let y₁, y₂,…,y_n be n solutions of the linear nth-order homogeneous DE on an interval I. Then, the set of solutions is linearly independent on I if and only if W(y₁, y₂,…,y_n) ≠ 0 for every x in the interval.

 Example: for y – 3y + 2y = 0, the two solutions y₁ = e^x and y₂ = e^2x has the Wronskian:

Wronskian Independence Checks

 The previous theorem implies that if y₁ and y₂ are two solutions of a linear homogeneous D.E., then

either W(y₁, y₂)  0 or W(y₁, y₂)  0, x.

 This can be proven by applying the existence and uniqueness theorem on zero initial condition IVP!

 For any two functions y₁ and y₂ that are not solutions of a linear homogeneous D.E. over an interval I:

 If W(y₁, y₂)  0, for some x I, then y₁ and y₂ are linearly independent over I.

 If W(y₁, y₂)  0, x, and y₁ & y₂ are nonzero with continuous derivatives in I^{, then y and y} are linearly dependent over I^.

Fundamental Set of Solutions

 Any set y₁, y₂, …, y_n of n linearly independent solutions of the homogeneous linear n^th-order DE on an interval I is said to be a fundamental set of solutions on the

interval I.

 Theorem: There exists a fundamental set of solutions for the homogeneous linear n^th-order DE.

 Similar to that a vector can be decomposed into linear combinations of basis vectors.

General Solution (1/2)

 Theorem: Let y₁, y₂,…,y_n be a fundamental set of

solutions of the homogeneous linear n^th-order DE on an interval I. Then, the general solution of the equation on the interval is

y = c₁y₁(x)＋c₂y₂(x)＋…＋c_ny_n(x), where c_i, i = 1,2,…,n are arbitrary constants.

Proof on n = 2:

Let Y be a solution of a₂(x)y" + a₁(x)y' + a₀(x)y = 0 on an interval I, y₁ and y₂ be linearly independent solutions of the DE. Initial conditions are Y(t) = k₁ and Y'(t) = k₂.

General Solution (2/2)

To solve for (c₁, c₂)^T, we have:

or

Since the Wronskian

given any k₁, k₂, there is always a unique solution

for c , c . #

Example: Linear Combo. of Solutions

 y₁ = e^3x and y₂ = e^–3x are both solutions of y" – 9y = 0 on the interval (–, ). Are they linearly independent? By observation? By Wronskian?

 Is y = 4sinh 3x – 5e^–3x a solution of y" – 9y = 0?

Nonhomogeneous Solutions (1/2)

 Theorem: Let y_p be any particular solution of the non- homogeneous linear n^th-order DE

on an interval I, and let y₁,y₂,…, y_n be a fundamental set of solutions. Then the general solution of the equation on the interval is:

y = c₁y₁(x)＋c₂y₂(x)＋…＋c_ny_n(x)＋y_p, where c , i = 1,2,…,n are arbitrary constants.

), ( )

( )

( ...

) ( )

( _{1 1 0}

1

1 a x y g x

dx x dy dx a

y x d

dx a y x d

a _n

n n n

n

n  _ ^_    

Nonhomogeneous Solutions (2/2)

Proof:

Let Y(x) and y_p(x) be particular solutions of L(y) = g(x).

Define u(x) = Y(x)－y_p(x), we have

L(u) = L{Y(x) – y_p(x)} = L(Y(x)) – L(y_p(x)) = g(x) – g(x) = 0 Thus, u(x) must be a solution to the homogeneous DE.

Therefore, u(x) = c₁y₁(x) + c₂y₂(x) +  + c_ny_n(x)

 Y(x) – y_p(x) = c₁y₁(x) + c₂y₂(x) +  + c_ny_n(x)

 Y(x) = c₁y₁(x) + c₂y₂(x) +  + c_ny_n(x) + y_p(x)

Any particular solution can be represented in this form.

#

Complementary Function

 The general solution of a homogeneous linear n^th-order DE is called the complementary function for the

associated non-homogeneous DE.

Let y_c(x) = c₁y₁(x)＋c₂y₂(x)＋…＋c_ny_n(x), the general

solution of a nonhomogeneous linear n^th-order DE has the form:

y(x) = y_c(x)＋y_p(x).

Superposition Principle for DE

 Theorem: Let y^p₁, y^p₂,…, y^p_k be k particular solutions of the non-homogeneous linear n^th-order DE on I,

corresponding to k distinct functions g₁, g₂,…,g_k. Then,

y^p = y^p1(x)＋ y^p2(x)＋…＋ y^pk(x) is a particular solution of

a_n(x)y⁽ⁿ⁾＋a_n–1(x)y^(n–1)＋…＋a₁(x)y'＋a₀(x)y

= g₁(x) + g₂(x) + …＋g_k(x).

Example of Superposition Principle

 Verify:

y^p₁ = –4x²  y" – 3y' + 4y = –16x² + 24x – 8 y^p₂ = e^2x  y" – 3y' + 4y = 2e^2x

y^p₃ = xe^x  y" – 3y' + 4y = 2xe^x – e^x Therefore

y = y^p₁ + y^p₂ + y^p₃ = –4x² + e^2x + xe^x is a solution of

y" – 3y' + 4y = –16x² + 24x – 8 + 2e^2x + 2xe^x – e^x

g₁(x) g₂(x) g₃(x)

Reduction of Order

 For a 2^nd order linear DE, one can construct a 2^nd

solution y₂ from a known nontrivial solution y₁. If y₁ and y₂ are linearly independent, we must have

y₂/y₁  constant,

Therefore, y₂(x) = u(x)y₁(x). Substitute this into the DE and solve for u(x) is called reduction of order.

Example: y" – y = 0, y

(x) = e

, find y

 Solution:

Given y₁(x) = e^x, let y₂(x) = u(x) e^x,

 y' = ue^x + e^xu', y" = ue^x + 2e^xu' + e^xu"

 y" – y = e^x(u"+ 2u') = 0

 u"+ 2u' = 0

Let w = u', the DE becomes w' + 2w = 0. Multiplying by the integrating factor e^2x, we have d[e^2xw]/dx = 0.

Therefore, w = c₁e^–2x or u' = c₁e^–2x.

 u = (–1/2) c₁e^–2x + c₂.

 y₂(x) = u(x) e^x = (–c₁/2) e^–x + c₂e^x, let c₁ = –2, c₂ = 0.

 Check W(e^x, e^–x)≠0

Solution by Reduction of Order (1/2)

 Put the 2^nd order DE into the standard form:

y" + P(x)y' + Q(x)y = 0,

where P(x) and Q(x) are continuous on some interval I.

If y₁ is a solution on I and that y₁(x) ≠ 0 for all x



y₂ + Py₂ + Qy₂ =

u[y₁+Py₁+Qy₁] + y₁u + (2y₁+Py₁)u = 0.

 y₁u + (2y₁ + Py₁)u = 0

Solution by Reduction of Order (2/2)

 Let w = u, we have y₁w + (2y₁ + Py₁)w = 0.

Since

. )

(

|

| ln

|

|

2 ln ₂

1 1

1 dx Pdx w y P x dx C

y y w

dw       







. )

(

|

|

ln ^wy₁²  



Example: x

y" – 3xy' + 4y = 0

 Since y₁ = x² is a known solution.



The general solution is y = c₁x²＋c₂x²lnx.

x x

x x dx

x e

e x dx

x e y

x y x y

y

x x dx x

dx

ln 4 0 3

2 2

3 / ln

3 4

/ 3 2

2

2

3







 

 













Constant Coefficients DE

 For homogeneous linear higher-order DE with real constant coefficients a_i, i = 0, 1, …, n, a_n  0, i.e.

a_ny⁽ⁿ⁾ + a_n-1y^(n–1) + … + a₂y" + a₁y' + a₀y = 0, do we have exponential solutions?

 Recall: by' + cy = 0,

y = c₁e^–ax on (–, ).

2 4 6 8 10 12 14 16

Auxiliary Equations

 Consider a 2nd-order DE, ay" + by' + cy = 0.

Let y = e^mx, and substituting y' = me^mx and y" = m²e^mx into the DE, we have: am²e^mx + bme^mx + ce^mx = 0.

e^mx > 0 for x  R  am² + bm + c = 0.

This is called the auxiliary equation of the DE.

General Solutions (1/2)

 Case I, b² – 4ac > 0:

m has two real roots m₁ and m₂, and y₁ = e^m1x and y₂ = e^m2x form a fundamental set of solutions.

The general solutions is

 Case II, b² – 4ac = 0:

m has one real root m₁ and y₁ = e^m1x. By reduction-of- order, the 2nd solution of the DE is y₂ = xe^m1x.

The general solution is

.

2

1 2

1

x m x

m c e

e c

y  

.

1

1 2

1

x m x

m c xe

e c

y  

General Solutions (2/2)

 Case III, b² – 4ac < 0:

m has two complex roots m₁ =









Similar to Case I, the general solution is:

 By proper selection of c₁ and c₂, and using Euler’s formula, e^iq = cos

q

q

) .

( 2 )

( 1

x i x

i c e

e c

y  ^{ }  ^{ }

).

sin cos

(c₁ x c₂ x e

y  ^x





Example: 4y"+4y'+17y = 0

 Solve the IVP: y(0) = –1, y'(0) = 2.

Solution:

The roots of the auxiliary equation 4m²+4m+17 = 0 are m₁ = – ½ + 2i and m₂ = – ½ – 2i

 y = e^–x/2 (c₁cos 2x + c₂sin 2x), with y(0) = –1, y'(0) = 2

 y = e^–x/2 (– cos 2x + ¾ sin 2x)

y  0, as x  .

x

1

y

1

Higher-Order Auxiliary Equations

 In general, to solve

a_ny⁽ⁿ⁾ + a_n–1y^(n–1) + … + a₂y" + a₁y' + a₀y = 0, where a_i  R and a_n  0, we must solve

a_nmⁿ + a_n–1m^n–1 + … + a₂m² + a₁m + a₀ = 0.

The general solution of the DE is:

Case I (no repeated roots):

Case II (with repeated roots):

.

... ¹

1

0 2

1

x m n x

m x

m c e c e _n

e c

y     ^

. ...

... ^{0 1}

0

0 1

1 2

1

x m n x

m k

x k m

k x

m x

m _{n k}

e c e

c e

x c e

x c e

c

y     ^  _   ^

Solution of Repeated Roots (1/2)

 For an n^th-order linear DE, assuming that the auxiliary equation of

a_ny⁽ⁿ⁾ + a_n–1y^(n–1) + … + a₁y' + a₀y = 0

has k repeated roots m₀. This means that the DE can be expressed as:

(D – m₀)^k(D – m₁) … (D – m_n–k)y = 0.

Hence, the solution of (D – m₀)^ky = 0 will also be a solution of the nth-order DE.

Solution of Repeated Roots (2/2)

 Since y₁ = e^m0x is a solution of (D – m₀)^ky = 0, let y(x) = u(x)e^m0x.

Note that

(D – m₀)[u(x)e^m0x] = (Du(x))e^m0x.

Applying the operator k times on y(x), we have (D – m₀)^k[u(x)e^m0x] = (D^ku(x))e^m0x for any u(x).

Then, u(x)e^m0x is a solution of the DE  D^ku(x) = 0.

Possible u(x) that meets this condition is a polynomial with degree less than k.

 y(x) = (c +c x+ … +c x^k–1)e^m0x is a family of solutions.

Non-homogeneous Linear DE

 To solve a non-homogeneous linear DE

a_ny⁽ⁿ⁾ + a_n-1y^(n–1) + … + a₂y" + a₁y' + a₀y = g(x), we must do two things:

(1) Find the complementary function y_c;

(2) Find any particular solution y_p of the DE.

Two methods:

 Method of undetermined coefficients

 Variation of parameters

Undetermined Coefficients (1/2)

 The method of undetermined coefficients can be applied under two conditions:

1. a_i, i = 0, 1, …, n, are constants, and

2. g(x) is a linear combination of functions of the following types:

P(x) = p_nxⁿ + p_n–1x^n–1 + … + p₂x² + p₁x + p₀, P(x)e^x,

P(x)e^xsin





Undetermined Coefficients (2/2)

 There are two approaches to find the particular solution given g(x) using the undetermined coefficients principle:

 Superposition approach (section 4.4 in the textbook)

 Assume that y_p(x) has similar form as g(x) with some coefficients to be determined

 Annihilator approach (section 4.5 in the textbook)

 Try to find a linear operator L_A such that when applied to both side of the DE turns it into a higher-order homogeneous DE. That is:

L(y) = g(x)  L_A  L(y) = L_A  g(x) = 0.

The extra solution subspace of L_A  L(y) = 0 should be the subspace of the particular solution.

Example: y" + 4y' – 2y = 2x

– 3x + 6

 By guessing, let y_p = Ax² + Bx + C, we have y_p' = 2Ax + B, and y_p" = 2A.

Therefore:

y_p" + 4y_p' – 2y_p

= 2A + 8Ax + 4B – 2Ax² – 2Bx – 2C

= – 2Ax² + (8A – 2B)x + (2A + 4B – 2C)

= 2x² – 3x + 6.

 y_p = – x² – (5/2)x – 9.

Example: y" – y' + y = 2 sin 3x

 By guessing, let y_p = A cos 3x + B sin 3x, we have

y_p' = – 3A sin 3x + 3B cos 3x, and y_p" = – 9A cos 3x – 9B sin 3x.

Therefore:

y_p" – y_p' + y_p

= (– 9A – 3B + A) cos 3x + (– 9B + 3A + B) sin 3x

= 2 sin 3x.

 y_p = (6/73) cos 3x – (16/73) sin 3x.

Example: y

by Superposition

 Solve y" – 2y' – 3y = 4x – 5 + 6xe^2x.

By super position principle, we divide the problem into two sub-problems, that is,

g(x) = g₁(x) + g₂(x),

where g₁(x) = 4x – 5, and g₂(x) = 6xe^2x.

By guessing, let y^p₁ = Ax + B, and y^p₂ = Cxe^2x + Ee^2x. Substitute y_p = Ax + B + Cxe^2x + Ee^2x into the DE, we have:

y_p = –(4/3)x + (23/9) – 2xe^2x – (4/3)e^2x

Example: A Glitch in the Method

 Solve y" – 5y' + 4y = 8e^x.

Simply guessing that y_p = Ae^x and substituting y_p into the DE gives us 0 = 8e^x. What went wrong?

If the guessed form of y_p falls in the solution space of y_c (i.e., y_c = c₁e^x + c₂e^4x), then we always get 0 = g(x).

Solution, let y_p = Axe^x. Since the derivatives of y_p

contains both the term Ae^x and Axe^x, it is a reasonable guess for a particular solution.

Summary of Two Cases (1/2)

 Case I:

No functions in the assumed particular solution is a solution of the associated homogeneous DE.

 Substitute with y_p = “the form of g(x)”.

g(x) y_p

1. 1 (any constant) 2. x³－x＋1

3. sin4x, or cos4x 4. e^5x

5. x²e^5x 6. e^3xsin4x 7. 5x²sin4x 8. xe^3xcos4x

A

Ax³＋Bx²＋Cx＋E A cos 4x＋B sin 4x Ae^5x

(Ax²＋Bx＋C)e^5x Ae^3xcos4x＋Be^3xsin4x

(Ax²＋Bx＋C)cos4x＋(Ex²＋Fx＋G)sin4x (Ax＋B)e^3xcos4x＋(Cx＋E)e^3xsin4x

Summary of Two Cases (2/2)

 Case II:

A function in the assumed particular solution is also a solution of the associated homogeneous DE.

 Substitute with y_p = xⁿ  “the form of g(x)”, where n is the smallest positive integer so that y_p is not in the

solution space of y_c.

Examples:

 Case I

 y" – 8y + 25y = 5x³e^–x – 7e^–x

 y" + 4y = x cos x

 y" – 9y + 14y = 3x² – 5 sin 2x + 7xe^6x

 Case II

 y" – 2y + y = e^x

 y" + y = 4x + 10 sin x, y() = 0, y() = 2

 y" – 6y + 9y = 6x² + 2 – 12 e^3x

Annihilator Approach

 The differential operators that annihilate different g(x) are as follows:

 Dⁿ annihilates 1, x, x², …, x^n–1.

 (D – )ⁿ annihilates e^x, xe^x, x²e^x, …, x^n–1e^x.

 [D² – 2D + (² + ²)]ⁿ annihilates e^xcosx, e^xsinx, xe^xcosx, xe^xsinx, … , x^n–1e^xcosx, x^n–1e^xsinx.

 Complementary solution to the annihilator DE gives you the form of y_p  you still need to substitute the solution form to determine the coefficients!

Example of Annihilator Approach

 Determine the y_p form of the DE: y + 3y + 2y = 4x². The annihilator of 4x² is D³. Thus, the root of the auxiliary equation of D³(y) = 0 is m = 0, 0, 0. The complementary solution is y = c₁ + c₂x + c₃x².

Therefore, the particular solution should have the form:

y_p = A + Bx + Cx².

 One advantage of the annihilator approach is that the y_c of L_A(y) = 0 and L(y) = 0 can be considered jointly to choose a y without glitch.

Variation of Parameters (1/3)

 To adopt the variation of parameters to a linear 2^nd- order DE a₂(x)y" + a₁(x)y + a₀(x)y = g(x), one must put the DE in the standard form:

y" + P(x)y + Q(x)y = f(x).

We seek a particular solution of the form y_p = u₁(x)y₁(x) + u₂(x)y₂(x),

where y₁ and y₂ form a fundamental set of solutions on I of the associated homogeneous DE.

Variation of Parameters (2/3)

 Take the derivatives y_p and y_p", and substitute them into the DE, we have

If y₁u₁^ + y₂u₂^ = h(x), then

   

 

     

 

 

) (

) (

) ( )

( )

( )

( ) ( )

(

2 2 1

1 2

2 1

1 2

2 1

1

2 2 1

1 2

2 1

1 2

2 1

1

2 2 1

1 2

2 1

1 2

2 2

2 1

1 1

1

2 2

2 2 1

1 1

1

x f u

y u

y u

y u

y x P u

y u

dx y d

u y u

y u

y u

y x P u

dx y u d

dx y d

u y u

y u

y u

y x P y

u u

y y

u u

y

y x Q y

x P y

u y

x Q y

x P y

u

y x Q y

x P

y_{p p p}

 

 



 

 

 

 

 



 



 

 

 

 

 



 



 

 

 



 

 

 

 

 

 



 





 

 





 



 

 





 

 

) ( ) ( )

( )

( ) (

2 2 1

1

2 2 1

1

x h x P x

h x

f u

y u

y

x h u

y u

y

Variation of Parameters (3/3)

 If we let h(x) = 0, then the solution of the system is

can be expressed in terms of determinants:

and where

W x f y W

u W_{1 2} ( )

1   





 

 





 

 

) ( 0

2 2 1

1

2 2 1

1

x f u

y u

y

u y u

y

) ,

1 (

2

2 W

x f y W

u  W 

) . ( , 0

) ( , 0

1 1 2

2 2 1

2 1

2 1

x f y

W y y

x f W y

y y

y W y

 

 



 

Summary of the Method

 To solve a₂(x)y" + a₁(x)y + a₀(x)y = g(x):

 Find y_c = c₁y₁ + c₂y₂.

 Compute the Wronskian W(y₁(x), y₂(x)).

 Put the DE into standard form: y" + P(x)y + Q(x)y = f(x).

 Find u₁ and u₂ by integrating u₁ = W₁/W and u₂ = W₂/W.

 A particular solution is y_p = u₁y₁ + u₂y₂.

 The general solution is y = y_c + y_p.

 Note that there is no need to introduce any constants when computing the indefinite integrals of u₁ and u₂.

Examples:

 Solve y" – 4y + 4y = (x + 1)e^2x.

 Solve 4y" + 36y = csc 3x.

 Solve y" – y = 1/x.

Higher-Order Equations

 For a linear nth-order DE

y⁽ⁿ⁾ + P_n-1(x)y^(n-1) + …+ P₁(x)y + P₀(x)y = f(x), if y_c = c₁y₁ + c₂y₂ + … + c_ny_n is the complementary function of the DE, then a particular solution is

y_p = u₁(x)y₁(x) + u₂(x)y₂(x) + … + u_n(x)y_n(x),

where u_k = W_k/W, k = 1, 2, …, n and W is the Wronskian of y₁, y₂, .., y_n and W_k is the determinant obtained by replacing the kth column of the Wronskian by the column (0, 0, …, f(x))^T.

Cauchy-Euler Equation

 Any linear differential equation of the form

where the coefficients a_i are constants, is called a Cauchy-Euler equation.

 Note that a_nxⁿ = 0 at x = 0. Therefore, we focus on solving the equation on (0, ).

),

0 (

1 1 1 1

1 a y g x

dx x dy dx a

y x d

dx a y x d

a _n

n n

n n n n

n  _ ^{ }_   

Method of Solution

 Assume that y = x^m is a solution, we have

. ) 1 )...(

2 )(

1 (

) 1 )...(

2 )(

1 (

...

) 1

( ²

2 2

1

m

k m k

k k k k k

m m

x k

m m

m m a

x k

m m

m m x a

dx y x d

a

x m

dx m y d dx mx dy



































2nd-Order Cauchy-Euler Eq.

 For the 2nd-order homogeneous equation:

a₂x²y + bxy + cy = 0, substituting y = x^m leads to

Thus y = x^m is a solution of the DE whenever m is a solution of the auxiliary equation

am(m – 1) + bm + c = 0.

. ) )

1 (

2 (

2

2 cy am m bm c xm

dx bx dy dx

y

ax d      

Auxiliary Equation Solutions (1/2)

 Case I, distinct real roots m₁≠m₂:

Then y₁ = x^m1 and y₂ = x^m2 form a fundamental set of solutions. The general solution is

 Case II, repeated real roots m₁ = m₂:

Then y₁ = x^m1, by reduction-of-order, the 2nd solution of the DE is y₂ = x^m1ln x. The general solution is

.

2

1 2

1

m

m c x

x c

y  

y .

Auxiliary Equation Solutions (2/2)

 Case III, conjugate complex roots:

If m₁ =









 By proper selection of c₁ and c₂, and using Euler’s formula, it can be shown that a general solution can also be represented by

).

( 2 )

( 1







 i c x i

x c

y  ^  ^

)).

ln sin(

) ln cos(

(c₁ x c₂ x

x

y  ^





Example: Particular Solutions

 The method of undetermined coefficients does not in general carry over to variable-coefficient DEs.

Therefore, the variation of parameters method should be used for solving non-homogeneous Cauchy-Euler equations.

 Example: Solve x²y – 3xy + 3y = 2x⁴e^x.

Reduction to Constant Coefficient Eqs

 A Cauchy-Euler equation can be reduced to a constant coefficient equation by the substitution x = e^t.

Note that dy/dt = dy/dxdx/dt = ye^t and d²y/dt² = ye^2t + ye^t. Thus, ax²y + bxy + cy = 0 can be reduced to

The constant coefficient technique can be used to solve y(t) and then y(x) in turn.

. 0 )

2 (

2 2

2 2

2      



 



 



 



 



 



   ^

 cy

dt a dy dt b

y a d

dt cy e dy be

e dt y

y e d

ae ^{t t t t t}

Nonlinear Equations

(1/2)

 Nonlinear DEs do not possess superposition property.

 For example, y₁ = e^x, y₂ = e^–x, y₃ = cos x, y₄ = sin x are four linearly independent solutions of the nonlinear 2^nd- order DE (y)² – y² = 0 on the interval (–, ). However, the following linear combinations are not solutions:

 y = c₁e^x + c₃cos x

 y = c₂e^–x + c₄sin x

 y = c₁e^x + c₂e^–x + c₃cos x + c₄sin x

Nonlinear Equations (2/2)

 We could find the one-parameter family of solutions of a few non-linear DEs, but these solutions are not

general solutions of the DEs.

 Higher order nonlinear DEs usually can not be solved analytically.

 Realistic physical models are often nonlinear.

Reduction of Order

 Nonlinear 2^nd-order DEs of the forms

 F(x, y, y) = 0

 F(y, y, y) = 0

can be reduced to 1^st-order DEs by letting u = y.

 For F(y, y, y) = 0, we have F(y, u, u) = 0.

 For F(y, y, y) = 0, observe that

So the problem becomes F(y, u, udu/dy) = 0.

dy . u du dx

dy dy du dx

y   du  

Example: y missing

 Solve y = 2x(y)² Solution:

Let u = y, du/dx = y, we have du/dx = 2xu²

 (1/u²) du = 2x dx   u^–2du =  2x dx

 – u^–1 = x² + c₁  –(y) ^–1 = x² + c₁

 dy/dx = –(x² + c₁) ^–1

 y = – (x² + c₁) ^–1 dx

. 1 tan

2 1

1 1

c c x

y   c 

 ^

Example: x missing

 Solve yy = (y)² Solution:

Let u = y, y = u du/dy, we have

 ln |u| = ln |y| + c₁  u = c₂y

  (1/y) dy = c₂ dx

 y = c₃e^c2x.

2 .

y dy u

u du dy

u du

y    



 





Example: Taylor Series Solution (1/2)

 Let us assume that a solution of the IVP exists:

y = x + y – y², y(0) = –1, y(0) = 1.

If y(x) is analytic at 0, we have the following Taylor series expansion centered at 0:

Note that

. 2 )

1 ( ) 1 ( 0 )

0 ( )

0 ( 0

) 0

(    ²      ²  

 y y

y



 

 

 

 ^{2 3}

! 3

) 0 (

! 2

) 0 (

! 1

) 0 ) (

0 ( )

( y x

y x y x

y x

y

Example: Taylor Series Solution (2/2)

For higher order derivatives, we have:

and so on.

Therefore, we have:

, 2

1 )

( )

( x y y² y yy

dx x d

y        

...

, ) ( 2 2

) 2

1 ( )

( ²

) 4

( y yy y yy y

dx x d

y        















 ^{2 3 4 5}

5 1 3

1 3

1 2 )

(x x x x x x

y