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Ordinary Differential Equations and Linear Algebra
1 Definition and Examples
We will move on to the second order equation which arises
a lot in science and
1.1 Definitions and First Example
Definition 1 A second-order differential equation
is an equation involving
the independent variable t and an unknown function y along with its first and
second derivatives. We will assume the second order derivative could be solved
explcitly. i.e. we will consider the equations of the form
y'' = f(t, y, y' )
A solution to the above equation is a twice continuously
y'' (t) ≡ f(t, y(t), y' (t))
Example 1 Newton’s law of mechanics involves
acceleration, which is second
order derivative of the position function. i.e The force may be
a function of the time t, the position x, and velocity dx/dt. So the second order
1.2 Linear Equation
We will focus on the Linear Equations . These
equation have the following
y'' + p(t)y' + q(t)y = g(t) (1)
As these are linear equations of y, y', y'' , we do not allow
1. the product of these to occur
2. nor any power higher than 1
3. nor any complicated function like sin y
Similar to the case of first-order linear equation, we can
consider the homogeneous
equation associated with (1)
y'' + p(t)y' + q(t)y = 0 (2)
The homogeneous equation plays an important role in the
solution of second
2 Existence and Uniqueness of Solutions:
Theorem 1 (Existence and Uniqueness) Suppose the
functions p(t), q(t), g(t)
are continuous on the interval ( α, β ). Let t0 be any point in ( α, β ). Then for
any real number , there is one and only one function y(t) defined on ( α, β )
which is a solution to
and satisfies the initial conditions and
Remark:: The major difference of the theorem here
from the existence and
uniqueness theorem we talked about yesterday is you can be assured the solution
exists wherever p, q, g is defined and continuous. So this is a global theorem.
3 Structure of the general solution:
Proposition 1 Suppose that
are both solutions to the homogeneous,
y'' + p(t)y' + q(t)y = 0 (3)
Then the function
is also a solution for any constants C1 and
C2. where we call y the linear combination of y1 and y2.
Example 2 For the simple harmonic motion equation
check x1(t) = cos
and x2 = sin
are both solutions to this equation. And so
is x(t) = C1x1(t) + C2x2(t).
It is easy to see x1(t) = sin
x2(t) = cos
are not constant multiples of
each other . Then we call them linearly independent.
Definition 2 Two functions u(t) and v(t) are said
to be linearly independent
on the interval ( α, β ) if neither is a constant multiple of each other over that
interval. If one is a constant multiple of the other on ( α, β ), they are said to be
linearly dependent there.
Example 3 check if the following pairs of functions
are linearly independent
1. x1 = sin t, x2 = sin t cos t
2. x1 = sin t, x2 = 0
3. x1 = , x2 =
4. x1 = sin t, x2 = sin t
For example, two solutions are linear dependent on each other, namely
x1(t) = Cx2(t). Then C1x1(t) + C2x2(t) = (C1C + C2)x2(t) = C0x2(t).
Now the natural question to ask is how we can tell whether
2 functions are
linearly independent or not. Some of them are not very easy to judge at the
We define the Wronskian of two functions u(t), v(t) to be
W(t) = u(t)v' (t) − v(t)u' (t)
he Wronskian could be used to tell if two solutions of the
second-order equation is linearly independent or not, due to the following
Proposition 2 Suppose the function u and v are
solutions to the linear, homogeneous
y'' + p(t)y' + q(t)y = 0 (4)
in the interval ( α, β ). Then u, v are linearly
independent if and only if the
Wronskian is identically zero .
Example 4 Exercise:
1. Compute the Wronskian of the solutions in example 2.
2. verify the solution of y'' +y' −6y = 0 is given by
and y1, y2 are linearly independent.
3. verify the solution of y'' −2y' +2y = 0 is given by
sin t and y1, y2 are linearly independent.