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
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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 .
Santanu Dey Lecture 6
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 .
Santanu Dey Lecture 6
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 .
Santanu Dey Lecture 6
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 .
Santanu Dey Lecture 6
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 .
Santanu Dey Lecture 6
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)
∣∣∣∣ .
Santanu Dey Lecture 6
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)
∣∣∣∣ .
Santanu Dey Lecture 6
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)
∣∣∣∣ .
Santanu Dey Lecture 6
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)
∣∣∣∣ .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 .
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 dependent
iff 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 .
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 .
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 .
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 .
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 .
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.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 .
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 .
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 .
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 .
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 .
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 .
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 .
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 .
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 .
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 .
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 .
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 .
Santanu Dey Lecture 6
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?
Santanu Dey Lecture 6
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?
Santanu Dey Lecture 6
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?
Santanu Dey Lecture 6
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?
Santanu Dey Lecture 6
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?
Santanu Dey Lecture 6
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?
Santanu Dey Lecture 6
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?
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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!)
Santanu Dey Lecture 6
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!)
Santanu Dey Lecture 6
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!)
Santanu Dey Lecture 6
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!)
Santanu Dey Lecture 6
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!)
Santanu Dey Lecture 6
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!)
Santanu Dey Lecture 6
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!)
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
Method of reduction of order
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.
Santanu Dey Lecture 6
Method of reduction of order
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.
Santanu Dey Lecture 6
Method of reduction of order
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.
Santanu Dey Lecture 6
Method of reduction of order
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.
Santanu Dey Lecture 6
Method of reduction of order
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.
Santanu Dey Lecture 6
Method of reduction of order
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.
Santanu Dey Lecture 6
Method of reduction of order
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.
Santanu Dey Lecture 6
Method of reduction of order
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.
Santanu Dey Lecture 6
Method of reduction of order
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.
Santanu Dey Lecture 6
Method of reduction of order
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.
Santanu Dey Lecture 6
Method of reduction of order
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.
Santanu Dey Lecture 6
Method of reduction of order
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.
Santanu Dey Lecture 6
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 .
Santanu Dey Lecture 6
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 .
Santanu Dey Lecture 6
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 .
Santanu Dey Lecture 6
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 .
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6
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.
Santanu Dey Lecture 6