Second Order Linear Differential equations

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Second Order Linear Differential equations

• In this lecture
• We define a second order Linear DE
• State the existence and uniqueness of solutions
  of second order Initial Value problems
• We find the general solution of a second order
  Homogeneous Linear DE
• Define the Wronskian of two functions and give a
  criterion for LI of two functions

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Definition Recall that a DE is said to be linear
if the dependent variable and its derivatives
occur only in the first degree.
Thus a second order linear DE in the standard
form is
. ()
where P(x), Q(x), and R(x) are continuous
functions of x on some interval a, b .
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Existence and uniqueness of solutions of a second
order Initial value problem
Theorem Let P(x), Q(x), and R(x) be continuous
functions of x on a closed interval a, b. Let
x0 be any point in the interval a, b and let
y0, y?0 be two real numbers. Then there exists a
unique solution y y(x) of the second order LDE
such that when x x0, y y0, y? y?0 .
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Problem (8, p 87) Show that y 0 and y x2sin
x are both solutions of
and that both satisfy the conditions y(0)0 and
y(0)0. Does this contradict Theorem A? If not,
why not?
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Henceforth whenever we will deal with a second
order LDE, we shall assume that the hypotheses of
Theorem A are satisfied.
That is to say, every equation we deal with will
have a solution.
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Second Order Homogeneous L.D.E.s
A second order homogeneous LDE is an equation of
the form
()
If y u(x) and y v(x) are two solutions of the
eqn. (), then any scalar linear combination y
? u(x) ? v(x) is also a solution of (). Here
?, ? are two real numbers.
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Follows from the fact that the derivative of sum
is sum of the derivatives, the derivative of
constant times a function is the same constant
times the derivative and that 0 0 0.
or
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Theorem If y yp(x) is a particular solution
of
. ()
and if y yh(x) is the general solution of the
associated homogeneous equation (), then the
general solution of () is
y yh(x) yp(x)
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Thus from the previous theorem, we find that to
solve a general second order LDE we have to find

• the general solution of the associated
  homogeneous equation
• a particular solution of the given second order
  linear differential equation
• and add the two to get the general solution of
  the given second order linear differential
  equation

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We let Ca, b as the set of all continuous real
valued functions on the interval a, b. We
define C?a, b as the set of all twice
continuously differentiable real valued functions
on the interval a, b. We know from Linear
Algebra that Ca, b and C?a, b are both real
vector spaces under pointwise addition and scalar
multiplication of functions. i.e. If f, g are two
functions, f g, a f are defined by (fg)(x)
f(x) g(x) (a f )(x) a f
(x) for all points x.
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The second order LDE
can be written in the compact form
where
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We thus have a Linear transformation
y
Ly
defined by
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The set of all solutions of the homogeneous
linear DE
equals
and hence is the null space of the linear
transformation L and hence is a subspace of
We now show that it is a
two-dimensional subspace of
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Theorem Let P(x), Q(x), and R(x) be continuous
functions of x on a closed interval a, b. The
set, S, of all solutions of the second order
homogeneous LDE
()
is a two-dimensional subspace of
Proof Fix a point x0 in the interval a,b. By
the existence and uniqueness theorem of a second
order Initial-value problem,
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there exists a unique solution y u(x) of ()
satisfying the initial conditions u(x0)
1, u? (x0) 0.
Again by the same theorem, there exists a unique
solution y v(x) of () satisfying the initial
conditions v(x0) 0, v? (x0) 1.
We show
(i) u(x), v(x) is LI.
(ii) u(x), v(x) span S.
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Let a u(x) ? v(x) 0 ..(1)
Proof of (i)
Differentiating (1) w.r.t. x, we get
a u?(x) ? v?(x) 0 ..(2) Put x x0 in
(1). We get a 1 ? 0 0 i.e. a 0 Put x
x0 in (2). We get a 0 ? 1 0 i.e. ?
0 Thus u(x), v(x) is LI.
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Proof of (ii)
Let y w(x) be a solution of the homogeneous
equation ().
Suppose w(x0) ?, w?(x0) ?.
Now y ? u(x) ? v(x) is also a solution of
() satisfying the initial conditions when x
x0, y ?, y? ?.
Hence by the uniqueness of the initial value
problem, w(x) ? u(x) ? v(x) or u(x), v(x)
span S.
Thus S has a basis consisting of two vectors.
Or S is two-dimensional.
Q.E.D.
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Example The 2nd order equation
(1)
Its solution space is a 2-dimensional subspace of
C(-?, ?).
It is easy to show that the functions
are solutions of (1) on (-?, ?), and since
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since the vectors (1,0) and (0, 1) are LI , coshx
and sinhx also form a basis for the solution
space of the equation. It follows that the
general solution of (1) may also be written
where c1 and c2 are arbt. Constants.
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The function
provide second pair of solutions of (1). In this
case
since the vectors (1,1) and (1, -1) are LI , ex
and e-x also form a basis for the solution space
of the equation. It follows that the general
solution of (1) may also be written
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Wronskian test for Linear Independence
Definition Let
be two
differentiable functions. Their Wronskian is
defined as the determinant
Note that the Wronskian is also a function of x.
We may also write
to show its dependence on
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Examples
1. Let
Their Wronskian is
1.
2. Let
Their Wronskian is
3. Let
Their Wronskian is
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Fact 1. Let
be two functions in
If their Wronskian is not zero
at some point x0 in a, b, then
is LI.
. (i)
Proof Let
Differentiating w.r.t. x, we get
. (ii)
Substituting x x0 in equations (i), (ii), we get
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These are two homogeneous linear equations in the
two unknowns c1, c2.
Determinant of the coefficient matrix is
Hence c1 0, c2 0.
or
is LI.
Q.E.D.
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Remark. The converse is not true.
is LI in
We can show that
But their Wronskian is zero.
We show that the converse is true if
are two solutions of a second
order homogeneous linear DE
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Abels formula for the Wronskian
Theorem Let
be two solutions of
of a second order homogeneous linear DE
()
Then their Wronskian is given by
for some constant C.
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Proof Since
are solutions of ()
(i)
(ii)
(i)y2 (ii) y1 gives
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is
The Wronskian of
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Thus we get
Separating the variables and integrating we get
C, an arbitrary constant
Corollary The Wronskian of two solutions of a
homogeneous second order linear DE is identically
zero or is never zero.
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Theorem Let the Wronskian of two
of a second order
solutions
homogeneous linear DE be zero. Then
is LD.
Proof Case (i) y1(x) is the zero function
Hence
is LD.
Case (ii) y1(x) is not zero
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0.
Hence
a constant.
or
i.e.
is LD.
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We now give an alternative proof.
Theorem Let the Wronskian of two solutions
of a second order homogeneous
linear DE be zero at some point x0 in a, b.
Then
is LD.
Proof Given
0.
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Consider the two homogeneous algebraic linear
equations in two unknowns x1, x2
The determinant of the coefficient matrix is
0.
Hence the above system has a nontrivial solution
(c1, c2). That is we can find c1, c2 (both not
zero , i.e. at least one of them ?0)
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such that
Now
is a solution of the homogeneous second order LDE
() such that y(x0) 0, y?(x0) 0.
But y 0 is also a solution of the homogeneous
second order LDE () such that y(x0) 0, y?(x0)
0. Hence by the uniqueness theorem,
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Thus we have found constants
c1, c2
(not both zero) such that
Thus
is LD.
Q.E.D.
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Examples 1.
are two solutions
of the second order linear d.e.
Find their Wronskian W(x).
By Abels formula,
an arbitrary constant
are two solutions of the
2.
second order linear d.e.
Their Wronskian W(x) is
C.
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Use of a known solution to find another.
Suppose y y1(x) is a known solution of the
second order LDE
()
We assume a second LI solution as
Now
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Substituting y y2 vy1 in (), we get
i.e.
This is zero
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i.e.
Let
Hence we get
Separating the variables we get
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Integrating, we get
Taking exponentials, we get
Again integrating, we get
We now show that
is LI, where
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is
The Wronskian of
? 0.
Hence
is LI.
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Thus a second LI solution of () is
where
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Condition satisfied
Solution
y ex
1P(x)Q(x)0
y e-x
1-P(x)Q(x)0
y eax
a2 aP(x)Q(x)0
a2 -aP(x)Q(x)0
y e-ax
y x
P(X) xQ(X)0
m(m -1)mxp(x)x2Q(x)0
y xm
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Examples 1. A solution of the d.e.
is
Find the complete solution.
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Solution
A second LI solution y2 is given by
where
Hence the complete solution is
where c1, c2 are arbitrary constants.
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Problem 9, page 94 Solve the DE
Solution Since the sum of the coefficients
is one solution.
Note P
Thus a second LI solution y2 is given by
where
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Hence the complete solution is
where c1, c2 are arbitrary constants.
(Note We have written c2/2 as a new constant c2.)
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Problem 5 page 94
The equation

• is the special case of Legenders equation

Corresponding to p 1, it has y1 x as an
obvious solution. Find the general solution.
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