Chapter 1 - Some General Theory of Ordinary Differential Equations

Question 1

Existence and Uniqueness Theorem

Theorem

Answer 1

-consider the differential equation
y’’ + p(x)y’ + q(x)y = 0
with p and q continuous on some interval I given by a≤x≤b
-let α, β be any two real numbers and xo be any point in the interval I
-then the equation has a unique solution defined on the interval I which satisfies:
y(xo) = α
y’(xo) = β

Question 2

Existence and Uniqueness Theorem

General Solution

Answer 2

y’’ + p(x)y’ + q(x)y = 0
-since the equation is linear and homogeneous, its solution set forms a vector space
-since it is second order, the dimension of this space is 2
-any two linearly independent solutions y1(x), y2(X) can be used as a basis
-the general solution is given as the linear superposition:
y(x) = c1 y1(x) + c2 y2(x)
-the existence and uniqueness theorem says that c1 and c2 can be fixed by specifying initial conditions

Question 3

y’’ + p(x)y’ + q(x)y = 0

Boundary Conditions

Answer 3

• it is also possible to fix c1 and c2 by specifying 2-point boundary conditions such as y(0) = y(1) = 0
• but in this case we do not have such a powerful general theorem telling us that a (nontrivial) solution always exists

Question 4

y’’ + p(x)y’ + q(x)y = 0

Why only two initial conditions?

Answer 4

-the equation implies that:
y''(x) = -p(x)y'(x) - q(x)y(x) 
for any x ∈ I 
-in particular:
y''(xo) = -p(xo)y'(xo) - q(xo)y(xo)
-so y''(xo) cannot be independently specified

Question 5

The Wronskian

Derivation

Answer 5

given:
y(x) = c1 y1(x) + c2 y2(x)
and
y(xo) = α , y’(xo) = β
-we have a pair of linear equations for c1 and c2:
c1 y1(xo) + c2y2(xo) = α
c1y1’(xo) + c2y2’(xo) =β
-rewrite in a matrix form:
AB = C
-where A = 2x2 matrix, top row y1(xo) , y2(xo), bottom row y1’(xo) , y2’(xo)
-B is a 2x1 matrix: c1, c2
-C is a 2x1 matrix: α, β
-to be able to calculate c1 and c2 we need to multiply both sides by A inverse, so A inverse must exist
-the Wronskian is the determinant of matrix A so if the Wronskian is 0 we cannot find c1 and c2

Question 6

Wronskian of Two Functions

Answer 6

W(x) = Wy1,y2

= y1(x)y2’(x) - y1’(x)y2(x)

Question 7

Abel’s Formula

Formula

Answer 7

-let yi(x) be solutions of :
y'' + p(x)y' + q(x)y = 0
Then:
W'(x) = -p(x) W(x)
and so:
W(x) = Wo*exp(-∫p(ξ)dξ)
-where the integral is taken from xo to x

Question 8

Abel’s Formula

Proof

Answer 8

-start with the definition of the Wronskian :
W(x) = y1y2’ - y1’y2
-differentiate:
W’(x) = y1’y2’ + y1y2’’ - y1’y2’ - y1’‘y2
W’(x) = y1y2’’ - y1’‘y2
-from the original ODE: y’’ + p(x)y’ + q(x)y = 0
we can write
y2’’ = -py2’ - qy2
y1’’ = -py1’ - qy1
-sub in and cancel:
W’(x) = -p(x)W(x)
-therefore defining Wo = W(xo), we have W’(x)=-p(x)W(x) solved by W(x) = Wexp(-∫p(ξ)dξ)
-where the integral is taken from xo to x

Question 9

Linear Dependence

Definition

Answer 9

-two functions y1(x) and y2(x) are linearly dependent if there exists γ≠0 such that y2 = γ*y1

Question 10

Wronskian of Linearly Dependent Functions

Answer 10

• if two functions y1 and y2 are linearly dependent then y2 can be written as y2 = γ*y1 for some non-zero γ
• in this case the Wronskian is 0

Question 11

Is this statement true?

‘ W[y1,y2] = 0 implies y1 and y2 are linearly dependent’

Answer 11

• no this is not always true and can be shown by counter example
• if y1=x^3 and y2=|x|^3 then y1 and y2 are linearly independent but W[y1,y2]=0
• however this statement does hold true subject to the condition that y1 and y2 are solutions of a second order homogeneous linear differential equation

Question 12

Linear Dependence and Wronskian Theorem

Answer 12

-let y1(x) and y2(x) be two non-zero solutions of:
y’’ + p(x)y’ + q(x)y = 0
-with p(x) and q(x) continuous on some interval I given by a≤x≤b
-then y1(x) and y2(x) are linearly dependent if and only if their Wronskian vanishes identically on I

Question 13

Euler’s Equation

Answer 13

-the general form of Euler's equation:
x²y'' + axy' + by = 0
-where a and b are constants
-in this case p(x)=ax/x² and q(x)=b/x²
-the point x=0 is badly defined so we seek solutions only for x>0
-calculating the Wronskian gives:
W(x) = Wo*x^(-a)

Question 14

Singular Point

Definition

Answer 14

-points which cannot be included in the interval I are called singular points
-e.g. if p(x)=1/x² and q(x)=1/x
then x=0 would be a singular point as the functions are not continuous over an interval including x=0

Question 15

Wronskian of Three Functions

Answer 15

• determinant of a 3x3 matrix
• first row with entries: y1, y2, y3
• second row with entries: y1’, y2’, y3’
• third row with entries: y1’’, y2’’, y3’’

Question 16

Wronskian of Three Functions to Find the Homogeneous Linear Differential Equation Whose Solution Space is Spanned by Two Given Functions

Answer 16

-two functions y1(x) and y2(x) are given as linearly independent solutions of:
y'' + p(x)*y' + q(x)*y = 0
-let y(x) be a general solution of this equation, then:
y(x) = c1*y1(x) + c2*y2(x) 
-for some c1 and c2
-calculate the Wronskian:
W[y1,y2,y] = 0 
W[y1,y2,y]=y'' + p(x)y' + q(x)y
-where:
p(x) = -W'(x)/W(x)
q(x) = W[y1',y2'] / W[y1,y2]

Question 17

y’’ + p(x)y’ + q(x)y = 0

Given One Solution, Find , the Other

Answer 17

-given one solution y1(x), then we have a first order differential equation for y2(x):
y1*y2' - y1'*y2 = W(X)
-this equation is linear in y2 with integrating factor y1^(-2) leading to:
d/dx (y2/y1) = W(x) / y1²
-integrating and rearranging:
y2 = y1 * ∫ W(x)/y1² dx
y2 = 
y1 *  ∫ [exp(- ∫ p(x)dx)/y1²]dx

Question 18

Constant Coefficient Second Order ODE with Double Root

y’’ - 2my’ + m²y = 0

Answer 18

y'' - 2my' + m²y = 0
-using y=e^(λx) gives an auxiliary equation:
(λ-m)² = 0
-so:
y1 = e^(mx)
-calculate W using 
W'=-p(x)W , in this case 
p(x)=-2m :
W' = 2mW
=> W = e^(2mx)
-substitute into 
d/dx (y2/y1) = W(x) / y1² :
y2 = x*e^(mx)

Question 19

Basis of Solutions

Answer 19

-to find the general solution of;
y’’ + p(x)y’ + q(x)y = 0
-we just need to find two linearly independent solutions
-i.e. solutions y1 and y2 such that W[y1,y2]≠0
-since all other solutions are then just a linear combination of these two, the pair of functions y1 and y2 form a basis in the linear algebra sense

Question 20

Standard Basis for The Solution Space of the Second Order Equation

Answer 20

y’’ + p(x)y’ + q(x)y = 0
-just like choosing a convenient basis e1 and e2 for 2-D space, we can choose:
->y1 to be the unique solution satisfying:
y1(xo) = 1
AND y1’(xo) = 0
->y2 to be the unique solution satisfying:
y2(xo) = 0
AND y2’(xo) = 1
-in this case W[y1,y2]=1 so y1 and y2 satisfy linear independence

Question 21

Standard Basis for the Solution Space of a Second Order Equation Proof

Answer 21

-for  the standard basi we choose:
y1(xo) = 1,  y1'(xo) = 0
AND
y2(xo) = 0, y2'(xo) = 1
-in this case 
W[y1,y2]=1≠0 
so y1 and y2 satisfy linear independence
-these conditions also satisfy the existence and uniqueness theorem:
let y(x) = αy1(x) + βy2(x)
-calculate y(xo):
y(xo) = αy1(xo) + βy2(x0)
= α*1  + 0 = α
-calculate y'(xo):
y'(xo) = αy1'(xo) + βy2'(x)
= 0 + β*1 = β

Question 22

Change of Basis

Answer 22

• let y1 and y2 be a basis
• we can form a new basis by non-singular linear transformation
• i.e. take the 2x1 column vector y1, y2 and multiply by the 2x2 vector a,b,c,d such that ad-bc≠0

Question 23

Change of Basis
Hyperbolic Function
y’’ - y = 0

Answer 23

y'' - y = 0
-this equation has solutions;
y1 = e^x
y2 = e^(-x)
-an alternative basis is:
^y1 = coshx
^y2 = sinhx
-these basis are related by the 2x2 matrix with entries
a=1/2 , b=1/2 , c=1/2 , d=-1/2

Question 24

Finding a Solution to Euler’s (Cauchy’s) Equation

Deriving the Indicial Equation

Answer 24

x²y'' + axy' + by = 0
-since:
y = x^r
=> x*y' = r*x^r
=> x²*y'' = r(r-a)*x^r
-we have:
x²y'' + axy' + by 
= [ r(r-1) + ar + b ] * x^r = 0
-if r is chosen to satisfy the indicial equation
F(r) = r² + (a-1)r + b = 0
-then y=x^r is a solution of Euler's Equation

Question 25

Finding a Solution to Euler’s (Cauchy’s) Equation

2 Distinct Real Roots

Answer 25

r = r1, r2
-in this case the general solution (for x>0) is :
y = c1x^(r1) + c2x^(r2)
-if r is complex or irrational then we define:
x^r = e^(r*logx)

Question 26

Finding a Solution to Euler’s (Cauchy’s) Equation

Double Root

Answer 26

(r-r1)² = 0
-we only have one solution, y1 = x^(r1)
-the solution is of the form:
y = (c1 + c2logx)x^(r1)

Question 27

Finding a Solution to Euler’s (Cauchy’s) Equation

Complex Conjugate Roots

Answer 27

r = α ± iβ
-writing:
x^r = e^((α ± iβ)*logx)
=e^(α*logx) [cos(βlogx)+isin(βlogx) ]]
-we have:
y1 = x^α cos(βlogx)
and
y2 = x^α sin(βlogx)
-as independent solutions

Question 28

Reduction of Order Equation

Answer 28

d/dx (y2/y1) = W(x) / y1²

Question 29

Euler’s Equation

Indicial Equation

Answer 29

x²y'' + axy' + by = 0
-using the substitution:
y= x^r * Σanx^n
=>
y' = Σ(n+r)*an*x^(n+r-1)
y'' = Σ(n+r-1)*(n+r)*an*x^(n+r-2)
-sub in to Euler's Equation
-choose n=0 to get the lowest possible power of x term: x^r
-equate the coefficient of x^r to 0, since ao≠0, the indicial equation is:
r² + (a-1)r + b = 0 = F(r)

Question 30

Defining a Function

Answer 30

• many functions arise in some specific context with no reference to differential equations e.g. trigonometric functions in geometry
• however it is possible to define well known functions as solutions of some differential equations and then derive some well known properties of these functions

Question 31

Properties of Solutions of y’ = y

Answer 31

-let E(x) be the solution of the initial value problem:
y’ = y , y(0)=1
-since the equation is autonomous, the function Er(x) = E(x+r) is also a solution
-since the equation is first order and linear there exists a constant a such that:
Er(x) = aE(x)
-we have Er(0)=aE(0), so a=E(r)
-we have shown that:
E(x+r) = E(x)E(r)
-one of the fundamental properties of the exponential function
-we can show that E(x)>0 for all x:
->suppose E(s)=0 for some real s, then for arbitrary x we have:
E(x+s) = E(x)E(s) = 0
so, E(x)=0, this contradicts E(0)=1

Question 32

Properties of the Solutions of y’‘+y=0

Answer 32

-if y(x) is any solution of 
y'' + y = 0
-then y' is also a solution since:
y'' + y' = 0 = (y')''+y'=0
-take S(x) as the unique solution satisfying S(0)=0 and S'(0)=1
-define C(x) = S'(x)
-then C'(x) = S''(x) = -S(x)
-since W' = p(x)*W and p(x)=0, W = constant
-calculate the Wronskian:
W[S,C] = -(S² + C²)
-specifically W for x=0 is -1
-since W is constant
- (S² + C²) = -1
S² + C² = 1
-we have derived a well known property of sine and cosine just from a differential equation

Question 33

Euler’s Equation

Solutions for Different Values of r

Answer 33

-where y is the general solution:
1) Two Distinct Real Roots:
y = c1x^(r1) + c2x^(r2)
2) Double Roots
(r-r1)²=0 , y1=x^(r1)
y = (c1+c2logx)x^r1
3) Complex Conjugate Roots
r=α±iβ
y = e^[(α±iβ)logx]
= e^(αlogx) * [cos(βlogx)+isin(βlogx)
=>
y1 = x^α * cos(βlogx)
y2 = x^α * sin(βlogx)