MA 108 - Ordinary Differential Equationsdey/diffeqn_autumn13/lecture6.pdf · If r(x) 0 in the equation d2y dx2 + p(x) dy dx + q(x)y = r(x); that is, d2y dx2 + p(x) dy dx + q(x)y =

MA 108 - Ordinary Differential Equations

Santanu Dey

Department of Mathematics,Indian Institute of Technology Bombay,

Powai, Mumbai [email protected]

October 9, 2013

Santanu Dey Lecture 6

Outline of the lecture

Second order linear equations

Method of reduction of order


Second order differential equations

Recall that a general second order linear ODE is of the form

a2(x)d2y

dx2+ a1(x)

dy

dx+ a0(x)y = g(x).

An ODE of the form

d2y

dx2+ p(x)

dy

dx+ q(x)y = r(x)

is called a second order linear ODE in standard form.Though there is no formula to find all the solutions of such anODE, we study the existence, uniqueness and number of solutionsof such ODE’s.




a2(x)d2y

dx2+ a1(x)

dy

dx+ a0(x)y = g(x).

An ODE of the form

d2y

dx2+ p(x)

dy

dx+ q(x)y = r(x)

is called a second order linear ODE in standard form.

Though there is no formula to find all the solutions of such anODE, we study the existence, uniqueness and number of solutionsof such ODE’s.




a2(x)d2y

dx2+ a1(x)

dy

dx+ a0(x)y = g(x).

An ODE of the form

d2y

dx2+ p(x)

dy

dx+ q(x)y = r(x)

is called a second order linear ODE in standard form.Though there is no formula to find all the solutions of such anODE, we study the existence, uniqueness and number of solutionsof such ODE’s.


Homogeneous Linear Second Order DE

If r(x) ≡ 0 in the equation

d2y

dx2+ p(x)

dy

dx+ q(x)y = r(x),

that is,d2y

dx2+ p(x)

dy

dx+ q(x)y = 0,

then the ODE is said to be homogeneous.Otherwise it is called non-homogeneous.




d2y

dx2+ p(x)

dy

dx+ q(x)y = r(x),

that is,d2y

dx2+ p(x)

dy

dx+ q(x)y = 0,





d2y

dx2+ p(x)

dy

dx+ q(x)y = r(x),

that is,d2y

dx2+ p(x)

dy

dx+ q(x)y = 0,

then the ODE is said to be homogeneous.

Otherwise it is called non-homogeneous.




d2y

dx2+ p(x)

dy

dx+ q(x)y = r(x),

that is,d2y

dx2+ p(x)

dy

dx+ q(x)y = 0,





d2y

dx2+ p(x)

dy

dx+ q(x)y = r(x),

that is,d2y

dx2+ p(x)

dy

dx+ q(x)y = 0,



Initial Value Problem- Existence/Uniqueness

An initial value problem of a second order homogeneous linearODE is of the form:

y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = a, y ′(x0) = b,

where p(x) and q(x) are assumed to be continuous on an openinterval I with x0 ∈ I , has a unique solution y(x) in the interval I .




y ′′ + p(x)y ′ + q(x)y = 0,

y(x0) = a, y ′(x0) = b,





y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = a,

y ′(x0) = b,





y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = a, y ′(x0) = b,





y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = a, y ′(x0) = b,



Linearly independent functions & Wronskian

Definition

The functions φ1(x) and φ2(x) are said to be linearly independenton an open interval I if

C1φ1(x) + C2φ2(x) = 0 ∀x ∈ I =⇒ C1 = C2 = 0.

Definition

Wronskian determinantThe Wronskian W (y1, y2) of two differentiable functions y1(x) andy2(x) is defined by

W (y1, y2)(x) :=

∣∣∣∣ y1(x) y2(x)y ′1(x) y ′2(x)

∣∣∣∣ .



Definition


C1φ1(x) + C2φ2(x) = 0 ∀x ∈ I =⇒ C1 = C2 = 0.

Definition


W (y1, y2)(x) :=

∣∣∣∣ y1(x) y2(x)y ′1(x) y ′2(x)

∣∣∣∣ .



Definition


C1φ1(x) + C2φ2(x) = 0 ∀x ∈ I =⇒ C1 = C2 = 0.

Definition


W (y1, y2)(x) :=

∣∣∣∣ y1(x) y2(x)y ′1(x) y ′2(x)

∣∣∣∣ .



Definition


C1φ1(x) + C2φ2(x) = 0 ∀x ∈ I =⇒ C1 = C2 = 0.

Definition


W (y1, y2)(x) :=

∣∣∣∣ y1(x) y2(x)y ′1(x) y ′2(x)

∣∣∣∣ .Santanu Dey Lecture 6

Test for l.i.

Suppose that

y ′′ + p(x)y ′ + q(x)y = 0

has continuous coefficients on an open interval I .

Then

1. two solutions y1 and y2 of the DE on I are linearly dependentiff their Wronskian is 0 at some x0 ∈ I .

2. Wronskian =0 for some x = x0 =⇒W ≡ 0 on I .

3. if there exists an x1 ∈ I at which W 6= 0, then y1 and y2 arelinearly independent on I .


Test for l.i.

Suppose that

y ′′ + p(x)y ′ + q(x)y = 0

has continuous coefficients on an open interval I . Then

1. two solutions y1 and y2 of the DE on I are linearly dependent

iff their Wronskian is 0 at some x0 ∈ I .




Test for l.i.

Suppose that

y ′′ + p(x)y ′ + q(x)y = 0






Test for l.i.

Suppose that

y ′′ + p(x)y ′ + q(x)y = 0



2. Wronskian =0 for some x = x0 =⇒

W ≡ 0 on I .



Test for l.i.

Suppose that

y ′′ + p(x)y ′ + q(x)y = 0






Test for l.i.

Suppose that

y ′′ + p(x)y ′ + q(x)y = 0






Proof

1. two solutions y1 and y2 of the DE on I are linearly dependent

iff theirWronskian is 0 at some x0 ∈ I .

Let y1, y2 be linearly dependent. Then, y1(x) = ky2(x), for someconstant k. Then, W (y1, y2) = W (ky2, y2) ≡ 0.Conversely, let W (y1, y2) = 0 for some x0 ∈ I . That is,

W (y1, y2)(x0) =

∣∣∣∣ y1(x0) y2(x0)y ′1(x0) y ′

2(x0)

∣∣∣∣ = 0.

Consider the linear system of equations :

k1y1(x0) + k2y2(x0) = 0

k1y′1(x0) + k2y

′2(x0) = 0

W (y1, y2)(x0) = 0 =⇒ ∃ non-trivial k1 & k2 solving the above linearsystem.Let y(x) = k1y1(x) + k2y2(x).Now, y(x0) = y ′(x0) = 0.

By existence-uniqueness theorem, y(x) ≡ 0 is the unique solution of

y ′′ + p(x)y ′ + q(x)y = 0; y(x0) = 0, y ′(x0) = 0 Thus

k1y1(x) + k2y2(x) = 0 with k1, k2 both not identically zero. Hence, y1, y2

are linearly dependent.


Proof

1. two solutions y1 and y2 of the DE on I are linearly dependent iff theirWronskian is 0 at some x0 ∈ I .

Let y1, y2 be linearly dependent.

Then, y1(x) = ky2(x), for someconstant k. Then, W (y1, y2) = W (ky2, y2) ≡ 0.Conversely, let W (y1, y2) = 0 for some x0 ∈ I . That is,

W (y1, y2)(x0) =

∣∣∣∣ y1(x0) y2(x0)y ′1(x0) y ′

2(x0)

∣∣∣∣ = 0.


k1y1(x0) + k2y2(x0) = 0

k1y′1(x0) + k2y

′2(x0) = 0







Proof


Let y1, y2 be linearly dependent. Then, y1(x) = ky2(x), for someconstant k .

Then, W (y1, y2) = W (ky2, y2) ≡ 0.Conversely, let W (y1, y2) = 0 for some x0 ∈ I . That is,

W (y1, y2)(x0) =

∣∣∣∣ y1(x0) y2(x0)y ′1(x0) y ′

2(x0)

∣∣∣∣ = 0.


k1y1(x0) + k2y2(x0) = 0

k1y′1(x0) + k2y

′2(x0) = 0







Proof


Let y1, y2 be linearly dependent. Then, y1(x) = ky2(x), for someconstant k . Then, W (y1, y2) =

W (ky2, y2) ≡ 0.Conversely, let W (y1, y2) = 0 for some x0 ∈ I . That is,

W (y1, y2)(x0) =

∣∣∣∣ y1(x0) y2(x0)y ′1(x0) y ′

2(x0)

∣∣∣∣ = 0.


k1y1(x0) + k2y2(x0) = 0

k1y′1(x0) + k2y

′2(x0) = 0







Proof


Let y1, y2 be linearly dependent. Then, y1(x) = ky2(x), for someconstant k . Then, W (y1, y2) = W (ky2, y2)

≡ 0.Conversely, let W (y1, y2) = 0 for some x0 ∈ I . That is,

W (y1, y2)(x0) =

∣∣∣∣ y1(x0) y2(x0)y ′1(x0) y ′

2(x0)

∣∣∣∣ = 0.


k1y1(x0) + k2y2(x0) = 0

k1y′1(x0) + k2y

′2(x0) = 0







Proof


Let y1, y2 be linearly dependent. Then, y1(x) = ky2(x), for someconstant k . Then, W (y1, y2) = W (ky2, y2) ≡ 0.

Conversely, let W (y1, y2) = 0 for some x0 ∈ I . That is,

W (y1, y2)(x0) =

∣∣∣∣ y1(x0) y2(x0)y ′1(x0) y ′

2(x0)

∣∣∣∣ = 0.


k1y1(x0) + k2y2(x0) = 0

k1y′1(x0) + k2y

′2(x0) = 0







Proof


Let y1, y2 be linearly dependent. Then, y1(x) = ky2(x), for someconstant k . Then, W (y1, y2) = W (ky2, y2) ≡ 0.Conversely, let W (y1, y2) = 0 for some x0 ∈ I .

That is,

W (y1, y2)(x0) =

∣∣∣∣ y1(x0) y2(x0)y ′1(x0) y ′

2(x0)

∣∣∣∣ = 0.


k1y1(x0) + k2y2(x0) = 0

k1y′1(x0) + k2y

′2(x0) = 0







Proof


Let y1, y2 be linearly dependent. Then, y1(x) = ky2(x), for someconstant k . Then, W (y1, y2) = W (ky2, y2) ≡ 0.Conversely, let W (y1, y2) = 0 for some x0 ∈ I . That is,

W (y1, y2)(x0) =

∣∣∣∣ y1(x0) y2(x0)y ′1(x0) y ′

2(x0)

∣∣∣∣ = 0.


k1y1(x0) + k2y2(x0) = 0

k1y′1(x0) + k2y

′2(x0) = 0







Proof



W (y1, y2)(x0) =

∣∣∣∣ y1(x0) y2(x0)y ′1(x0) y ′

2(x0)

∣∣∣∣ = 0.


k1y1(x0) + k2y2(x0) = 0

k1y′1(x0) + k2y

′2(x0) = 0







Proof



W (y1, y2)(x0) =

∣∣∣∣ y1(x0) y2(x0)y ′1(x0) y ′

2(x0)

∣∣∣∣ = 0.


k1y1(x0) + k2y2(x0) = 0

k1y′1(x0) + k2y

′2(x0) = 0







Proof



W (y1, y2)(x0) =

∣∣∣∣ y1(x0) y2(x0)y ′1(x0) y ′

2(x0)

∣∣∣∣ = 0.


k1y1(x0) + k2y2(x0) = 0

k1y′1(x0) + k2y

′2(x0) = 0







Proof



W (y1, y2)(x0) =

∣∣∣∣ y1(x0) y2(x0)y ′1(x0) y ′

2(x0)

∣∣∣∣ = 0.


k1y1(x0) + k2y2(x0) = 0

k1y′1(x0) + k2y

′2(x0) = 0







Proof



W (y1, y2)(x0) =

∣∣∣∣ y1(x0) y2(x0)y ′1(x0) y ′

2(x0)

∣∣∣∣ = 0.


k1y1(x0) + k2y2(x0) = 0

k1y′1(x0) + k2y

′2(x0) = 0

W (y1, y2)(x0) = 0 =⇒ ∃ non-trivial k1 & k2 solving the above linearsystem.

Let y(x) = k1y1(x) + k2y2(x).Now, y(x0) = y ′(x0) = 0.






Proof



W (y1, y2)(x0) =

∣∣∣∣ y1(x0) y2(x0)y ′1(x0) y ′

2(x0)

∣∣∣∣ = 0.


k1y1(x0) + k2y2(x0) = 0

k1y′1(x0) + k2y

′2(x0) = 0

W (y1, y2)(x0) = 0 =⇒ ∃ non-trivial k1 & k2 solving the above linearsystem.Let y(x) = k1y1(x) + k2y2(x).

Now, y(x0) = y ′(x0) = 0.






Proof



W (y1, y2)(x0) =

∣∣∣∣ y1(x0) y2(x0)y ′1(x0) y ′

2(x0)

∣∣∣∣ = 0.


k1y1(x0) + k2y2(x0) = 0

k1y′1(x0) + k2y

′2(x0) = 0







Proof



W (y1, y2)(x0) =

∣∣∣∣ y1(x0) y2(x0)y ′1(x0) y ′

2(x0)

∣∣∣∣ = 0.


k1y1(x0) + k2y2(x0) = 0

k1y′1(x0) + k2y

′2(x0) = 0



y ′′ + p(x)y ′ + q(x)y = 0;

y(x0) = 0, y ′(x0) = 0 Thus




Proof



W (y1, y2)(x0) =

∣∣∣∣ y1(x0) y2(x0)y ′1(x0) y ′

2(x0)

∣∣∣∣ = 0.


k1y1(x0) + k2y2(x0) = 0

k1y′1(x0) + k2y

′2(x0) = 0



y ′′ + p(x)y ′ + q(x)y = 0; y(x0) = 0,

y ′(x0) = 0 Thus




Proof



W (y1, y2)(x0) =

∣∣∣∣ y1(x0) y2(x0)y ′1(x0) y ′

2(x0)

∣∣∣∣ = 0.


k1y1(x0) + k2y2(x0) = 0

k1y′1(x0) + k2y

′2(x0) = 0



y ′′ + p(x)y ′ + q(x)y = 0; y(x0) = 0, y ′(x0) = 0

Thus




Proof



W (y1, y2)(x0) =

∣∣∣∣ y1(x0) y2(x0)y ′1(x0) y ′

2(x0)

∣∣∣∣ = 0.


k1y1(x0) + k2y2(x0) = 0

k1y′1(x0) + k2y

′2(x0) = 0




k1y1(x) + k2y2(x) = 0 with k1, k2 both not identically zero.

Hence, y1, y2



Proof



W (y1, y2)(x0) =

∣∣∣∣ y1(x0) y2(x0)y ′1(x0) y ′

2(x0)

∣∣∣∣ = 0.


k1y1(x0) + k2y2(x0) = 0

k1y′1(x0) + k2y

′2(x0) = 0





are linearly dependent.Santanu Dey Lecture 6

Proof contd..

2. Wronskian =0 for some x = x0 =⇒

W ≡ 0 on I .

If Wronskian =0 for some x = x0, then by the first part of theresult, y1 & y2 are linearly dependent=⇒ W (y1, y2)(x) = 0 ∀x ∈ I .

3. if there exists an x1 ∈ I at which W 6= 0, then y1 and y2 are l.i. on I .

W (y1, y2)(x1) 6= 0 =⇒ y1 & y2 can’t be linearly dependent=⇒ y1 & y2 are l.i.

Definition

A basis or fundamental set of solutions of y ′′ + p(x)y ′ + q(x)y = 0on an interval I is a pair y1, y2 of linearly independent solutions ofy ′′ + p(x)y ′ + q(x)y = 0 on I .


Proof contd..





Definition



Proof contd..


If Wronskian =0 for some x = x0,

then by the first part of theresult, y1 & y2 are linearly dependent=⇒ W (y1, y2)(x) = 0 ∀x ∈ I .



Definition



Proof contd..


If Wronskian =0 for some x = x0, then by the first part of theresult, y1 & y2 are linearly dependent

=⇒ W (y1, y2)(x) = 0 ∀x ∈ I .



Definition



Proof contd..





Definition



Proof contd..





Definition



Proof contd..




W (y1, y2)(x1) 6= 0

=⇒ y1 & y2 can’t be linearly dependent=⇒ y1 & y2 are l.i.

Definition



Proof contd..




W (y1, y2)(x1) 6= 0 =⇒ y1 & y2 can’t be linearly dependent

=⇒ y1 & y2 are l.i.

Definition



Proof contd..




W (y1, y2)(x1) 6= 0 =⇒ y1 & y2 can’t be linearly dependent

=⇒ y1 & y2 are l.i.

Definition



Proof contd..





Definition



Proof contd..





Definition



Proof contd..





Definition



Examples

1. The continuity of p(x) and q(x) is required in the results of theprevious slide.

Consider the DE

x2y ′′ − 4xy ′ + 6y = 0.

Then, x2 and x3 are linearly independent solutions, butW (x2, x3)(0) = 0.2. Consider y1(x) = x2 and

y2(x) =

{x2 if x ≥ 0

−x2 if x < 0,

Then, W (y1, y2)(x) = 0 for all x ∈ R, but y1 and y2 are linearlyindependent. Does it contradict the result in the previous slide?


Examples

1. The continuity of p(x) and q(x) is required in the results of theprevious slide. Consider the DE

x2y ′′ − 4xy ′ + 6y = 0.


y2(x) =

{x2 if x ≥ 0

−x2 if x < 0,



Examples


x2y ′′ − 4xy ′ + 6y = 0.

Then, x2 and x3 are linearly independent solutions, butW (x2, x3)(0) = 0.

2. Consider y1(x) = x2 and

y2(x) =

{x2 if x ≥ 0

−x2 if x < 0,



Examples


x2y ′′ − 4xy ′ + 6y = 0.

Then, x2 and x3 are linearly independent solutions, butW (x2, x3)(0) = 0.2. Consider y1(x) = x2

and

y2(x) =

{x2 if x ≥ 0

−x2 if x < 0,



Examples


x2y ′′ − 4xy ′ + 6y = 0.


y2(x) =

{x2 if x ≥ 0

−x2 if x < 0,



Examples


x2y ′′ − 4xy ′ + 6y = 0.


y2(x) =

{x2 if x ≥ 0

−x2 if x < 0,

Then, W (y1, y2)(x) = 0 for all x ∈ R,

but y1 and y2 are linearlyindependent. Does it contradict the result in the previous slide?


Examples


x2y ′′ − 4xy ′ + 6y = 0.


y2(x) =

{x2 if x ≥ 0

−x2 if x < 0,



Basis of solutions

Result : If p(x) and q(x) are continuous on an open interval I , theny ′′ + p(x)y ′ + q(x)y = 0 has a basis of solutions on I .

Proof : Consider the IVP’s

y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = 1, y ′(x0) = 0

y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = 0, y ′(x0) = 1

By existence-uniqueness theorem of IVP, the above problems haveunique solutions y1(x) and y2(x) respectively on I .

Now, W (y1, y2)(x0) =

∣∣∣∣ 1 00 1

∣∣∣∣ = 1 6= 0 =⇒ y1 &y2 are l.i. Why?

Hence, they form a basis of solutions of y ′′ + p(x)y ′ + q(x)y = 0.


Basis of solutions


Proof :

Consider the IVP’s

y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = 1, y ′(x0) = 0

y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = 0, y ′(x0) = 1


Now, W (y1, y2)(x0) =

∣∣∣∣ 1 00 1

∣∣∣∣ = 1 6= 0 =⇒ y1 &y2 are l.i. Why?



Basis of solutions



y ′′ + p(x)y ′ + q(x)y = 0,

y(x0) = 1, y ′(x0) = 0

y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = 0, y ′(x0) = 1


Now, W (y1, y2)(x0) =

∣∣∣∣ 1 00 1

∣∣∣∣ = 1 6= 0 =⇒ y1 &y2 are l.i. Why?



Basis of solutions



y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = 1, y ′(x0) = 0

y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = 0, y ′(x0) = 1


Now, W (y1, y2)(x0) =

∣∣∣∣ 1 00 1

∣∣∣∣ = 1 6= 0 =⇒ y1 &y2 are l.i. Why?



Basis of solutions



y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = 1, y ′(x0) = 0

y ′′ + p(x)y ′ + q(x)y = 0,

y(x0) = 0, y ′(x0) = 1


Now, W (y1, y2)(x0) =

∣∣∣∣ 1 00 1

∣∣∣∣ = 1 6= 0 =⇒ y1 &y2 are l.i. Why?



Basis of solutions



y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = 1, y ′(x0) = 0

y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = 0, y ′(x0) = 1


Now, W (y1, y2)(x0) =

∣∣∣∣ 1 00 1

∣∣∣∣ = 1 6= 0 =⇒ y1 &y2 are l.i. Why?



Basis of solutions



y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = 1, y ′(x0) = 0

y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = 0, y ′(x0) = 1


Now, W (y1, y2)(x0) =

∣∣∣∣ 1 00 1

∣∣∣∣ = 1 6= 0 =⇒ y1 &y2 are l.i. Why?



Basis of solutions



y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = 1, y ′(x0) = 0

y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = 0, y ′(x0) = 1


Now, W (y1, y2)(x0) =

∣∣∣∣ 1 00 1

∣∣∣∣ = 1 6= 0 =⇒ y1 &y2 are l.i. Why?



Basis of solutions



y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = 1, y ′(x0) = 0

y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = 0, y ′(x0) = 1


Now, W (y1, y2)(x0) =

∣∣∣∣ 1 00 1

∣∣∣∣

= 1 6= 0 =⇒ y1 &y2 are l.i. Why?



Basis of solutions



y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = 1, y ′(x0) = 0

y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = 0, y ′(x0) = 1


Now, W (y1, y2)(x0) =

∣∣∣∣ 1 00 1

∣∣∣∣ = 1 6= 0

=⇒ y1 &y2 are l.i. Why?



Basis of solutions



y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = 1, y ′(x0) = 0

y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = 0, y ′(x0) = 1


Now, W (y1, y2)(x0) =

∣∣∣∣ 1 00 1

∣∣∣∣ = 1 6= 0 =⇒ y1 &y2 are l.i.

Why?



Basis of solutions



y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = 1, y ′(x0) = 0

y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = 0, y ′(x0) = 1


Now, W (y1, y2)(x0) =

∣∣∣∣ 1 00 1

∣∣∣∣ = 1 6= 0 =⇒ y1 &y2 are l.i. Why?



General solution

Let y1 & y2 be a basis of solutions of the homogeneous second orderlinear DE y ′′ + p(x)y ′ + q(x)y = 0 on I , where p(x) and q(x) arecontinuous on I .

Then,

y(x) = C1y1(x) + C2y2(x)

is a general solution of y ′′ + p(x)y ′ + q(x)y = 0.That is, every solution y = Y (x) of the DE has the form

Y (x) = C1y1(x) + C2y2(x),

where C1 and C2 are arbitrary constants.


General solution

Let y1 & y2 be a basis of solutions of the homogeneous second orderlinear DE y ′′ + p(x)y ′ + q(x)y = 0 on I , where p(x) and q(x) arecontinuous on I . Then,

y(x) = C1y1(x) + C2y2(x)

is a general solution of y ′′ + p(x)y ′ + q(x)y = 0.

That is, every solution y = Y (x) of the DE has the form

Y (x) = C1y1(x) + C2y2(x),



General solution


y(x) = C1y1(x) + C2y2(x)


Y (x) = C1y1(x) + C2y2(x),



General solution


y(x) = C1y1(x) + C2y2(x)


Y (x) = C1y1(x) + C2y2(x),



General solution


y(x) = C1y1(x) + C2y2(x)


Y (x) = C1y1(x) + C2y2(x),



Proof

Let Y (x) be a solution of the given ODE.

We want to find C1 andC2 such that

Y (x) = C1y1(x) + C2y2(x).

This implies for x0 ∈ I ,

Y (x0) = C1y1(x0) + C2y2(x0)

Y ′(x0) = C1y′1(x0) + C2y

′2(x0).

Thus, (y1(x0) y2(x0)y ′1(x0) y ′2(x0)

)[C1

C2

]=

[Y (x0)Y ′(x0)

].

As y1 and y2 form a basis of solutions of the DE,

W (x0) =

(y1(x0) y2(x0)y ′1(x0) y ′2(x0)

)is invertible. (Justify!)


Proof

Let Y (x) be a solution of the given ODE. We want to find C1 andC2 such that

Y (x) = C1y1(x) + C2y2(x).


Y (x0) = C1y1(x0) + C2y2(x0)

Y ′(x0) = C1y′1(x0) + C2y

′2(x0).

Thus, (y1(x0) y2(x0)y ′1(x0) y ′2(x0)

)[C1

C2

]=

[Y (x0)Y ′(x0)

].


W (x0) =

(y1(x0) y2(x0)y ′1(x0) y ′2(x0)



Proof


Y (x) = C1y1(x) + C2y2(x).


Y (x0) = C1y1(x0) + C2y2(x0)

Y ′(x0) = C1y′1(x0) + C2y

′2(x0).

Thus, (y1(x0) y2(x0)y ′1(x0) y ′2(x0)

)[C1

C2

]=

[Y (x0)Y ′(x0)

].


W (x0) =

(y1(x0) y2(x0)y ′1(x0) y ′2(x0)



Proof


Y (x) = C1y1(x) + C2y2(x).


Y (x0) = C1y1(x0) + C2y2(x0)

Y ′(x0) = C1y′1(x0) + C2y

′2(x0).

Thus, (y1(x0) y2(x0)y ′1(x0) y ′2(x0)

)[C1

C2

]=

[Y (x0)Y ′(x0)

].


W (x0) =

(y1(x0) y2(x0)y ′1(x0) y ′2(x0)



Proof


Y (x) = C1y1(x) + C2y2(x).


Y (x0) = C1y1(x0) + C2y2(x0)

Y ′(x0) = C1y′1(x0) + C2y

′2(x0).

Thus, (y1(x0) y2(x0)y ′1(x0) y ′2(x0)

)[C1

C2

]=

[Y (x0)Y ′(x0)

].


W (x0) =

(y1(x0) y2(x0)y ′1(x0) y ′2(x0)



Proof


Y (x) = C1y1(x) + C2y2(x).


Y (x0) = C1y1(x0) + C2y2(x0)

Y ′(x0) = C1y′1(x0) + C2y

′2(x0).

Thus, (y1(x0) y2(x0)y ′1(x0) y ′2(x0)

)[C1

C2

]=

[Y (x0)Y ′(x0)

].


W (x0) =

(y1(x0) y2(x0)y ′1(x0) y ′2(x0)



Proof


Y (x) = C1y1(x) + C2y2(x).


Y (x0) = C1y1(x0) + C2y2(x0)

Y ′(x0) = C1y′1(x0) + C2y

′2(x0).

Thus, (y1(x0) y2(x0)y ′1(x0) y ′2(x0)

)[C1

C2

]=

[Y (x0)Y ′(x0)

].


W (x0) =

(y1(x0) y2(x0)y ′1(x0) y ′2(x0)



Proof contd...

Therefore,

C1 =

∣∣∣∣ Y (x0) y2(x0)Y ′(x0) y ′2(x0)

∣∣∣∣det W (x0)

,

and

C2 =

∣∣∣∣ y1(x0) Y (x0)y ′1(x0) Y ′(x0)

∣∣∣∣det W (x0)

.

(Is the representation of Y in terms of C1 and C2 unique?)Now,

u(x) = Y (x)− C1y1(x)− C2y2(x)

satisfies the given DE, and

u(x0) = 0 = u′(x0).

But the constant function u(x) ≡ 0 also satisfies the IVP. Thus,

Y (x) = C1y1(x) + C2y2(x) by the uniqueness theorem.


Proof contd...

Therefore,

C1 =

∣∣∣∣ Y (x0) y2(x0)Y ′(x0) y ′2(x0)

∣∣∣∣det W (x0)

,

and

C2 =

∣∣∣∣ y1(x0) Y (x0)y ′1(x0) Y ′(x0)

∣∣∣∣det W (x0)

.


u(x) = Y (x)− C1y1(x)− C2y2(x)


u(x0) = 0 = u′(x0).




Proof contd...

Therefore,

C1 =

∣∣∣∣ Y (x0) y2(x0)Y ′(x0) y ′2(x0)

∣∣∣∣det W (x0)

,

and

C2 =

∣∣∣∣ y1(x0) Y (x0)y ′1(x0) Y ′(x0)

∣∣∣∣det W (x0)

.


u(x) = Y (x)− C1y1(x)− C2y2(x)


u(x0) = 0 = u′(x0).




Proof contd...

Therefore,

C1 =

∣∣∣∣ Y (x0) y2(x0)Y ′(x0) y ′2(x0)

∣∣∣∣det W (x0)

,

and

C2 =

∣∣∣∣ y1(x0) Y (x0)y ′1(x0) Y ′(x0)

∣∣∣∣det W (x0)

.


u(x) = Y (x)− C1y1(x)− C2y2(x)

satisfies the given DE,

and

u(x0) = 0 = u′(x0).




Proof contd...

Therefore,

C1 =

∣∣∣∣ Y (x0) y2(x0)Y ′(x0) y ′2(x0)

∣∣∣∣det W (x0)

,

and

C2 =

∣∣∣∣ y1(x0) Y (x0)y ′1(x0) Y ′(x0)

∣∣∣∣det W (x0)

.


u(x) = Y (x)− C1y1(x)− C2y2(x)


u(x0) = 0 = u′(x0).




Proof contd...

Therefore,

C1 =

∣∣∣∣ Y (x0) y2(x0)Y ′(x0) y ′2(x0)

∣∣∣∣det W (x0)

,

and

C2 =

∣∣∣∣ y1(x0) Y (x0)y ′1(x0) Y ′(x0)

∣∣∣∣det W (x0)

.


u(x) = Y (x)− C1y1(x)− C2y2(x)


u(x0) = 0 = u′(x0).

But the constant function u(x) ≡ 0

also satisfies the IVP. Thus,



Proof contd...

Therefore,

C1 =

∣∣∣∣ Y (x0) y2(x0)Y ′(x0) y ′2(x0)

∣∣∣∣det W (x0)

,

and

C2 =

∣∣∣∣ y1(x0) Y (x0)y ′1(x0) Y ′(x0)

∣∣∣∣det W (x0)

.


u(x) = Y (x)− C1y1(x)− C2y2(x)


u(x0) = 0 = u′(x0).

But the constant function u(x) ≡ 0 also satisfies the IVP.

Thus,



Proof contd...

Therefore,

C1 =

∣∣∣∣ Y (x0) y2(x0)Y ′(x0) y ′2(x0)

∣∣∣∣det W (x0)

,

and

C2 =

∣∣∣∣ y1(x0) Y (x0)y ′1(x0) Y ′(x0)

∣∣∣∣det W (x0)

.


u(x) = Y (x)− C1y1(x)− C2y2(x)


u(x0) = 0 = u′(x0).




Proof contd...

Therefore,

C1 =

∣∣∣∣ Y (x0) y2(x0)Y ′(x0) y ′2(x0)

∣∣∣∣det W (x0)

,

and

C2 =

∣∣∣∣ y1(x0) Y (x0)y ′1(x0) Y ′(x0)

∣∣∣∣det W (x0)

.


u(x) = Y (x)− C1y1(x)− C2y2(x)


u(x0) = 0 = u′(x0).





We’ve been looking at the second order linear homogeneous ODE

y ′′ + p(x)y ′ + q(x)y = 0.

As we remarked earlier, there is no general method to find a basisof solutions. However, if we know one non-zero solution y1(x) thenwe have a method to find y2(x) such that y1(x) and y2(x) arelinearly independent.To find such a y2(x), set

y2(x) = v(x)y1(x)

We’ll choose v such that y1 and y2 are linearly independent.Can v be a constant? No.Now for y2 to be a solution of the given ODE

y ′′2 + p(x)y ′2 + q(x)y2 = 0.

that is,(vy1)′′ + p(x)(vy1)′ + q(x)(vy1) = 0.




y ′′ + p(x)y ′ + q(x)y = 0.

As we remarked earlier, there is no general method to find a basisof solutions.

However, if we know one non-zero solution y1(x) thenwe have a method to find y2(x) such that y1(x) and y2(x) arelinearly independent.To find such a y2(x), set

y2(x) = v(x)y1(x)


y ′′2 + p(x)y ′2 + q(x)y2 = 0.





y ′′ + p(x)y ′ + q(x)y = 0.

As we remarked earlier, there is no general method to find a basisof solutions. However, if we know one non-zero solution y1(x) thenwe have a method to find y2(x) such that y1(x) and y2(x) arelinearly independent.

To find such a y2(x), set

y2(x) = v(x)y1(x)


y ′′2 + p(x)y ′2 + q(x)y2 = 0.





y ′′ + p(x)y ′ + q(x)y = 0.

As we remarked earlier, there is no general method to find a basisof solutions. However, if we know one non-zero solution y1(x) thenwe have a method to find y2(x) such that y1(x) and y2(x) arelinearly independent.To find such a y2(x),

set

y2(x) = v(x)y1(x)


y ′′2 + p(x)y ′2 + q(x)y2 = 0.





y ′′ + p(x)y ′ + q(x)y = 0.


y2(x) = v(x)y1(x)


y ′′2 + p(x)y ′2 + q(x)y2 = 0.





y ′′ + p(x)y ′ + q(x)y = 0.


y2(x) = v(x)y1(x)

We’ll choose v such that

y1 and y2 are linearly independent.Can v be a constant? No.Now for y2 to be a solution of the given ODE

y ′′2 + p(x)y ′2 + q(x)y2 = 0.





y ′′ + p(x)y ′ + q(x)y = 0.


y2(x) = v(x)y1(x)

We’ll choose v such that y1 and y2 are linearly independent.

Can v be a constant? No.Now for y2 to be a solution of the given ODE

y ′′2 + p(x)y ′2 + q(x)y2 = 0.





y ′′ + p(x)y ′ + q(x)y = 0.


y2(x) = v(x)y1(x)

We’ll choose v such that y1 and y2 are linearly independent.Can v be a constant?

No.Now for y2 to be a solution of the given ODE

y ′′2 + p(x)y ′2 + q(x)y2 = 0.





y ′′ + p(x)y ′ + q(x)y = 0.


y2(x) = v(x)y1(x)

We’ll choose v such that y1 and y2 are linearly independent.Can v be a constant? No.

Now for y2 to be a solution of the given ODE

y ′′2 + p(x)y ′2 + q(x)y2 = 0.





y ′′ + p(x)y ′ + q(x)y = 0.


y2(x) = v(x)y1(x)


y ′′2 + p(x)y ′2 + q(x)y2 = 0.





y ′′ + p(x)y ′ + q(x)y = 0.


y2(x) = v(x)y1(x)


y ′′2 + p(x)y ′2 + q(x)y2 = 0.





y ′′ + p(x)y ′ + q(x)y = 0.


y2(x) = v(x)y1(x)


y ′′2 + p(x)y ′2 + q(x)y2 = 0.



Second solution

Thus,

0 = (v ′y1 + vy ′1)′ + p(v ′y1 + vy ′1) + qvy1

= v ′′y1 + 2v ′y ′1 + vy ′′1 + p(v ′y1 + vy ′1) + qvy1

= v(y ′′1 + py ′1 + qy1) + v ′(2y ′1 + py1) + v ′′y1.

Thus,v ′′

v ′= −2y ′1 + py1

y1= −2y ′1

y1− p.

Therefore,

ln |v ′| = ln

(1

y21

)−∫

pdx ;

That is,

v =

∫e−

Rpdx

y21

dx .


Second solution

Thus,

0 = (v ′y1 + vy ′1)′ + p(v ′y1 + vy ′1) + qvy1

= v ′′y1 + 2v ′y ′1 + vy ′′1 + p(v ′y1 + vy ′1) + qvy1

= v(y ′′1 + py ′1 + qy1) + v ′(2y ′1 + py1) + v ′′y1.

Thus,v ′′

v ′= −2y ′1 + py1

y1= −2y ′1

y1− p.

Therefore,

ln |v ′| = ln

(1

y21

)−∫

pdx ;

That is,

v =

∫e−

Rpdx

y21

dx .


Second solution

Thus,

0 = (v ′y1 + vy ′1)′ + p(v ′y1 + vy ′1) + qvy1

= v ′′y1 + 2v ′y ′1 + vy ′′1 + p(v ′y1 + vy ′1) + qvy1

= v(y ′′1 + py ′1 + qy1) + v ′(2y ′1 + py1) + v ′′y1.

Thus,v ′′

v ′= −2y ′1 + py1

y1= −2y ′1

y1− p.

Therefore,

ln |v ′| = ln

(1

y21

)−∫

pdx ;

That is,

v =

∫e−

Rpdx

y21

dx .


Second solution

Thus,

0 = (v ′y1 + vy ′1)′ + p(v ′y1 + vy ′1) + qvy1

= v ′′y1 + 2v ′y ′1 + vy ′′1 + p(v ′y1 + vy ′1) + qvy1

= v(y ′′1 + py ′1 + qy1) + v ′(2y ′1 + py1) + v ′′y1.

Thus,v ′′

v ′= −2y ′1 + py1

y1= −2y ′1

y1− p.

Therefore,

ln |v ′| = ln

(1

y21

)−∫

pdx ;

That is,

v =

∫e−

Rpdx

y21

dx .


Second solution

Claim:

y1 and vy1 are linearly independent.Proof. Enough to check Wronskian!

W (y1, vy1) = y1(v ′y1 + y ′1v)− y ′1vy1

= y21 v ′

= y21

e−R

pdx

y21

= e−R

pdx

6= 0.


Second solution

Claim: y1 and vy1 are linearly independent.

Proof. Enough to check Wronskian!

W (y1, vy1) = y1(v ′y1 + y ′1v)− y ′1vy1

= y21 v ′

= y21

e−R

pdx

y21

= e−R

pdx

6= 0.


Second solution

Claim: y1 and vy1 are linearly independent.Proof.

Enough to check Wronskian!

W (y1, vy1) = y1(v ′y1 + y ′1v)− y ′1vy1

= y21 v ′

= y21

e−R

pdx

y21

= e−R

pdx

6= 0.


Second solution

Claim: y1 and vy1 are linearly independent.Proof. Enough to check Wronskian!

W (y1, vy1) = y1(v ′y1 + y ′1v)− y ′1vy1

= y21 v ′

= y21

e−R

pdx

y21

= e−R

pdx

6= 0.


Second solution

Claim: y1 and vy1 are linearly independent.Proof. Enough to check Wronskian!

W (y1, vy1) = y1(v ′y1 + y ′1v)− y ′1vy1

= y21 v ′

= y21

e−R

pdx

y21

= e−R

pdx

6= 0.


Documents

MA 108 - Ordinary Differential Equationsdey/diffeqn_autumn13/lecture6.pdf · If r(x) 0 in the equation d2y dx2 + p(x) dy dx + q(x)y = r(x); that is, d2y dx2 + p(x) dy dx + q(x)y =