Second Order Differential Equations
Ordinary Differential Equations
Dr. Marco A Roque Sol
12/01/2015
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Solutions of Linear Homogeneus Systems
Solutions of Linear Homogeneus Systems. The Wrosnkian
Solutions of Linear Homogeneus Systems
If y1 (t) and y2 (t) are two solutions to a linear, second order
homogeneous differential equation and they are smooth enough,
then the general solution to the linear, second order differential
equation is given by the equation given by
y (t) = c1 y1 (t) + c2 y2 (t)
The next question that we can ask is how to find the constants c1
and c2 . Since we have two constants it makes sense, hopefully,
that we will need two equations, or conditions, to find them.
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Solutions of Linear Homogeneus Systems
Solutions of Linear Homogeneus Systems. The Wrosnkian
Solutions of Linear Homogeneus Systems
One way to do this is to specify the value of the solution at two
distinct points, or,
y (t0 ) = y0
and
y (t1 ) = y1
These are typically called boundary values and are not really the
focus of this course.
Another way to find the constants would be to specify the value of
the solution and its derivative at a particular point
y (t0 ) = y0
and
y 0 (t0 ) = y00
These are the two conditions that we will be using here.
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Solutions of Linear Homogeneus Systems
Solutions of Linear Homogeneus Systems. The Wrosnkian
Solutions of Linear Homogeneus Systems
Example 31
Solve the following IVP.
y 00 − 9y = 0;
y (0) = 2, y 0 (0) = −1
Solution
The general solution to our differential equation is then
y (t) = c1 e −3t + c2 e 3t
Now all we need to do is apply the initial conditions. This means
that we need the derivative of the solution.
y 0 (t) = −3c1 e −3t + 3c2 e −3t
Plug in the initial conditions
2 = y (0) = c1 + c2
and
Dr. Marco A Roque Sol
− 1 = y 0 (0) = −3c1 + 3c2
Ordinary Differential Equations
Second Order Differential Equations
Solutions of Linear Homogeneus Systems
Solutions of Linear Homogeneus Systems. The Wrosnkian
Solutions of Linear Homogeneus Systems
This gives us a system of two equations and two unknowns that
can be solved. Doing this yields
7
6
The solution to the IVP is then,
c1 =
c2 =
5
6
5
7
y (t) = e −3t + e 3t
6
6
What conclusions can we draw from the preceding example that
will help us to deal with the more general equation
ay 00 + by 0 + cy = 0
,
whose coefficients a, b, and c are arbitrary (real) constants?
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Solutions of Linear Homogeneus Systems
Solutions of Linear Homogeneus Systems. The Wrosnkian
Solutions of Linear Homogeneus Systems
In the first place, in the example the solutions were exponential
functions. Further, once we had identified two solutions, we were
able to use a linear combination of them to satisfy the given initial
conditions as well as the differential equation itself.
It turns out that we can start by seeking exponential solutions of
the form y = e rt , where r is a parameter to be determined. Then
it follows that y 0 = re rt and y 00 = r 2 e rt . By substituting these
expressions for y , y 0 , and y 00 in the diferential equation, we obtain
(ar 2 + br + c)e rt = 0,
or, since e rt 6= 0,
ar 2 + br + c = 0,
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Solutions of Linear Homogeneus Systems
Solutions of Linear Homogeneus Systems. The Wrosnkian
Solutions of Linear Homogeneus Systems
The above equation,is called the characteristic equation for the
differential equation ay 00 + by 0 + cy = 0.
This means that if r is a root of the polynomial equation, then
y = e rt is a solution of the differential equation. Since the
characteristic equation is a quadratc one with real coefficients, it
has two roots, which may be real and different, real but repeated,
or complex conjugates. We consider the first case here.
Assuming that the roots of the characteristic equation are real and
different, let them be denoted by r1 and r2 , where r1 6= r2 . Then
y1 (t) = e r1 t and y2 (t) = e r2 t are two solutions of differential
equation. It follows that
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Solutions of Linear Homogeneus Systems
Solutions of Linear Homogeneus Systems. The Wrosnkian
Solutions of Linear Homogeneus Systems
y 0 (t) = r1 c1 e r1 t + r2 c2 e r2 t
and
y 00 (t) = r12 c1 e r1 t + r22 c2 e r2 t
Substituting these expressions for y , y 0 , and y 00 in the differential
equation
ay 00 + by 0 + cy = a(r12 c1 e r1 t + r22 c2 e r2 t ) + b(r1 c1 e r1 t + r2 c2 e r2 t ) + ...
... + c(c1 e r1 t + c2 e r2 t )
and rearranging terms, we obtain
ay 00 + by 0 + cy = c1 (ar12 c1 + br1 + c)e r1 t + c2 (ar22 c1 + br2 + c)e r2 t
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Solutions of Linear Homogeneus Systems
Solutions of Linear Homogeneus Systems. The Wrosnkian
Solutions of Linear Homogeneus Systems
The quantities in the two sets of parentheses on the right-hand
side are zero because r1 and r2 are roots of the characteristic
equation; therefore, y as given by
y (t) = c1 e r1 t + c2 e r2 t
is indeed a solution of the homogeneus equation, as we wished to
verify.
Now suppose that we want to find the particular member of the
family of solutions that satisfies the initial conditions
y (t0 ) = y0 ;
Dr. Marco A Roque Sol
y 0 (t0 ) = y00
Ordinary Differential Equations
Second Order Differential Equations
Solutions of Linear Homogeneus Systems
Solutions of Linear Homogeneus Systems. The Wrosnkian
Solutions of Linear Homogeneus Systems
By substituting t = t0 , y = y0 , and y 0 = y00 in the function and its
derivative, we obtain
c1 e r1 t0 + c2 e r2 t0 = y0
r1 c1 e r1 t0 + r2 c2 e r2 t0 = y00
On solving the above system for c1 and c2 , we find that
c1 =
y00 − y0 r2 −r1 t0
e
r1 − r2
c2 =
y0 r1 − y00 −r2 t0
e
r1 − r2
Recall that r1 − r2 6= 0 so that the expressions in the above
equations, always make sense.
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Solutions of Linear Homogeneus Systems
Solutions of Linear Homogeneus Systems. The Wrosnkian
Solutions of Linear Homogeneus Systems
Example 32
Find the general solution of
y 00 + 5y 0 + 6y = 0
The characteristic equation is
r 2 + 5r + 6 = (r + 2)(r + 3) = 0
Thus the possible values of r are r1 = 2 and r2 = 3; the general
solution of the equation is
y = c1 e −2t + c2 e −3t
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Solutions of Linear Homogeneus Systems
Solutions of Linear Homogeneus Systems. The Wrosnkian
Solutions of Linear Homogeneus Systems
Example 33
Find the general solution to the IVP
y 00 + 5y 0 + 6y = 0 y (0) = 2, y 0 (0) = 2
The general solution of the differential equation was found in the
previous example. Applying initiail condition on the function we get
c1 + c2 = 2
And applying initial conditions on y 0 = 2c1 e −2t − 3c2 e −3t , we
obtain
−2c1 − 3c2 = 3
By solving the system for c1 and c2 , we have c1 = 9 and c2 = −7
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Solutions of Linear Homogeneus Systems
Solutions of Linear Homogeneus Systems. The Wrosnkian
Solutions of Linear Homogeneus Systems
Using these values, we obtain the solution
y = 9e −2t − 7e −3t
Example 34
Find the general solution to the IVP
y 00 + 11y 0 + 24y = 0;
y (0) = 0, y 0 (0) = 7
The characteristic equation is
r 2 + 11r + 24 = 0
Its roots are r1 = −8 and r2 = −3 and so the general solution and
its derivative are.
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Solutions of Linear Homogeneus Systems
Solutions of Linear Homogeneus Systems. The Wrosnkian
Solutions of Linear Homogeneus Systems
y = c1 e −8t + c2 e −3t
y 0 = −8c1 e −8t − 3c2 e −3t
And applying initial conditions on the function and its derivative,
we obtain
0 = y (0) = c1 + c2
7 = y 0 (0) = −8c1 − 3c2
By solving the system for c1 and c2 , we have
c1 =
7
5
and
Dr. Marco A Roque Sol
c2 = −
7
5
Ordinary Differential Equations
Solutions of Linear Homogeneus Systems
Solutions of Linear Homogeneus Systems. The Wrosnkian
Second Order Differential Equations
Solutions of Linear Homogeneus Systems
Using these values, we obtain the solution
7
7
y = e −8t − e −3t
5
5
Example 35
Find the general solution to the IVP
4y 00 − 8y 0 + 3y = 0;
y (0) = 2, y 0 (0) =
1
2
The characteristic equation is
4r 2 − 8r + 3 = 0
Its roots are r1 =
derivative are.
3
2
and r2 =
3
1
2
1
y = c1 e 2 t + c2 e 2 t ;
Dr. Marco A Roque Sol
and so the general solution and its
3
1
3
1
y 0 = c1 e 2 t + c2 e 2 t
2
2
Ordinary Differential Equations
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And applying initial conditions on the function and its derivative,
we obtain
2 = y (0) = c1 + c2
1
3
1
= y 0 (0) = c1 + c2
2
2
2
By solving the system for c1 and c2 , we have
1
5
and c2 =
2
2
Using these values, we obtain the solution
c1 = −
1 3
5 1
y = − e 2t + e 2t
2
2
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Solutions of Linear Homogeneus Systems
Solutions of Linear Homogeneus Systems. The Wrosnkian
Solutions of Linear Homogeneus Systems. The
Wrosnkian
In the preceding section we showed how to solve some differential
equations of the form
ay 00 + by 0 + cy = 0,
where a, b, and c are constants. To discuss general properties of
linear differential equations, it is helpful to introduce a differential
operator notation. Let p and q be continuous functions on an open
interval I -that is, for α < t < β. The cases for α = −∞, or
β = ∞, or both, are included. Then, for any function φ that is
twice differentiable on I, we define the differential operator L by
the equation
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Solutions of Linear Homogeneus Systems
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Wrosnkian
L[φ] = φ00 + pφ0 + qφ,
Note that L[φ] is a function on I. The value of L[φ] at a point t is
L[φ](t) = φ00 (t) + p(t)φ0 (t) + q(t)φ(t),
For example, if p(t) = t 2 , q(t) = 1 + t, and φ(t) = sin3t, then
L[φ](t) = (sin3t)00 + t 2 (sin3t)0 + (1 + t)(sin3t)
L[φ](t) = −9sin3t + 3t 2 cost3t + (1 + t)(sin3t)
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Solutions of Linear Homogeneus Systems
Solutions of Linear Homogeneus Systems. The Wrosnkian
Solutions of Linear Homogeneus Systems. The
Wrosnkian
The operator L is often written as L = D 2 + pD + q, where D is
the derivative operator. In this section we study the second order
linear homogeneous equation L[φ](t) = 0. Since it is customary to
use the symbol y to denote φ(t), we will usually write this
equation in the form
L[y ](t) = y 00 (t) + p(t)y 0 (t) + q(t)y (t) = 0
With the above equation we associate a set of initial conditions
y (t0 ) = y0
y 0 (t0 ) = y00
where t0 is any point in the interval I, and y0 and y00 are given real
numbers. The fundamental theoretical result for initial value
problems for second order linear equations is stated below
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
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Theorem (Existence and Uniqueness Theorem)
Consider the initial value problem
L[y ](t) = y 00 (t) + p(t)y 0 (t) + q(t)y (t) = g (t);
y (t0 ) = y0 , y 0 (t0 ) = y00
where p, q, and g are continuous on an open interval I that
contains the point t0 . Then there is exactly one solution y = φ(t)
of this problem, and the solution exists throughout the interval I.
We emphasize that the theorem says three things:
1. The initial value problem has a solution; in other words, a
solution exists.
2. The initial value problem has only one solution; that is, the
solution is unique.
3. The solution φ is defined throughout the interval I where the
coefficients are continuous and is at least twice differentiable there.
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Solutions of Linear Homogeneus Systems
Solutions of Linear Homogeneus Systems. The Wrosnkian
Solutions of Linear Homogeneus Systems. The
Wrosnkian
For some problems some of these assertions are easy to prove. For
instance, we can find that the initial value problem
L[y ](t) = y 00 − y = 0;
y (0) = 2,
y 0 (0) = −1
has the solution
3
1
y (t) = e t + e −t
2
2
For most problems of the general form,
y 00 (t) + p(t)y 0 (t) + q(t)y (t) = g (t)
it is not possible to write down a useful expression for the solution.
Therefore, all parts of the theorem must be proved by general
methods.
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Solutions of Linear Homogeneus Systems
Solutions of Linear Homogeneus Systems. The Wrosnkian
Solutions of Linear Homogeneus Systems. The
Wrosnkian
Example 36
Find the longest interval in which the solution of the initial value
problem
(t 2 − 3t)y 00 (t) + ty 0 (t) − (t + 3)y (t) = 0;
y (1) = 2, y 0 (1) = 1
is certain to exist.
Solution
If the given differential equation is written in the form of
y 00 (t) + p(t)y 0 (t) + q(t)y (t) = g (t)
then p(t) = t/(t 2 − 3t), q(t) = −(t + 3)/(t 2 − 3t), and g (t) = 0.
The only points of discontinuity of the coefficients are t = 0 and
t = 3.
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
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Thus, the longest interval in which the Theorem guarantees that
the solution exists, is the longest open interval, containing the
initial point t = 1, in which all the coefficients are continuous is
0 < t < 3.
Example 37
Find the unique solution of the initial value problem
y 00 (t) + p(t)y 0 (t) + q(t)y (t) = 0;
y (t0 ) = 0, y 0 (t0 ) = 0
where p and q are continuous in an open interval I containing t0 .
Solution
We can see inmediately that the function y = 0 for all t is clearly a
solution for the IVP. Now, by the uniqueness part of the Theorem
mentioned above, it is the only solution of the given problem.
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Solutions of Linear Homogeneus Systems
Solutions of Linear Homogeneus Systems. The Wrosnkian
Solutions of Linear Homogeneus Systems. The
Wrosnkian
Now, recall from
(Principle of Superposition)
If y1 and y2 are two solutions of the differential equation
L[y ] = y 00 (t) + p(t)y 0 (t) + q(t)y (t) = 0
then the linear combination c1 y1 + c2 y2 is also a solution for any
values of the constants c1 and c2 .
It is easy to prove this result
L[c1 y1 + c2 y2 ] = (c1 y1 + c2 y2 )00 (t) + p(t)(c1 y1 + c2 y2 )0 + ...
... + q(t)(c1 y1 + c2 y2 )
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Solutions of Linear Homogeneus Systems
Solutions of Linear Homogeneus Systems. The Wrosnkian
Solutions of Linear Homogeneus Systems. The
Wrosnkian
L[c1 y1 + c2 y2 ] = c1 y100 + c2 y200 (t) + c1 p(t)y10 + c2 p(t)y20 + ...
... + c1 q(t)y1 + c2 q(t)y2
L[c1 y1 + c2 y2 ] = c1 y100 + c1 p(t)y10 + c1 q(t)y1 + ...
... + c2 y200 (t) + c2 p(t)y20 + c2 q(t)y2
L[c1 y1 + c2 y2 ] = c1 (y100 + p(t)y10 + q(t)y1 ) + ...
... + c2 (y200 (t) + c2 p(t)y20 + q(t)y2 )
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Solutions of Linear Homogeneus Systems
Solutions of Linear Homogeneus Systems. The Wrosnkian
Solutions of Linear Homogeneus Systems. The
Wrosnkian
L[c1 y1 + c2 y2 ] = c1 L[y1 ] + c2 L[y2 ]
Since L[y1 ] = 0 and L[y2 ] = 0, it follows that L[c1 y1 + c2 y2 ] = 0
also.
The next question is uniqueness. We begin to address this question
by examining whether the constants c1 and c2 can be chosen so as
to satisfy the initial conditions . These initial conditions require c1
and c2 to satisfy the equations
c1 y1 (t0 ) + c2 y2 (t0 ) = y0
c1 y10 (t0 ) + c2 y20 (t0 ) = y00
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Solutions of Linear Homogeneus Systems
Solutions of Linear Homogeneus Systems. The Wrosnkian
Solutions of Linear Homogeneus Systems. The
Wrosnkian
The determinant of the coefficients of the system is
y1 (t0 ) y2 (t0 )
= y1 (t0 )y20 (t0 ) − y10 (t0 )y2 (t0 )
W = 0
y1 (t0 ) y20 (t0 )
If W 6= 0, then the above equations have a unique solution (c1 , c2 )
regardless of the values of y0 and y00 . This solution is given by
c1 =
y0 y20 (t0) − y00 y2 (t0)
y1 (t0 )y20 (t0) − y10 (t0 )y2 (t0)
c2 =
−y0 y10 (t0) + y00 y1 (t0)
y1 (t0 )y20 (t0) − y10 (t0 )y2 (t0)
or in determinants
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Solutions of Linear Homogeneus Systems
Solutions of Linear Homogeneus Systems. The Wrosnkian
Solutions of Linear Homogeneus Systems. The
Wrosnkian
y0 y2 (t0 )
y 0 y 0 (t0 )
0
2
c1 = y1 (t0 ) y2 (t0 )
y 0 (t0 ) y 0 (t0 )
1
2
y1 (t0 ) y0 y 0 (t0 ) y 0 1
0
c2 = y1 (t0 ) y2 (t0 )
y 0 (t0 ) y 0 (t0 )
1
2
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Solutions of Linear Homogeneus Systems
Solutions of Linear Homogeneus Systems. The Wrosnkian
Solutions of Linear Homogeneus Systems. The
Wrosnkian
With these values for c1 and c2 , the linear combination
y = c1 y1 (t) + c2 y2 (t) satisfies the initial conditions as well as the
differential equation.
On the other hand, if W = 0, then the denominator appearing in
the above equations is zero. In this case the equations for c1 and
c2 have no solution unless y0 and y00 have values that also make
the numerators on those equations equal to zero.
The determinant W is called the Wronskian.
(Wronskian determinants are named for Jozef Maria
Hoene-Wronski (17761853), who was born in Poland but spent
most of his life in France, or simply the Wronskian, of the solutions
y1 and y2 .
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Solutions of Linear Homogeneus Systems
Solutions of Linear Homogeneus Systems. The Wrosnkian
Solutions of Linear Homogeneus Systems. The
Wrosnkian
Sometimes we use the more extended notation W (y1 , y2 )(t0 ),
emphasizing that the Wronskian depends on the functions y1 and
y2 , and that it is evaluated at the point t0 .
Theorem
Suppose that y1 and y2 are two solutions of equation
L[y ] = y 00 (t) + p(t)y 0 (t) + q(t)y (t) = 0
and that the initial conditions
y (t0 ) = y0 ,
y 0 (t0 ) = y00
are assigned. Then it is always possible to choose the constants
c1 , c2 so that
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
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Wrosnkian
y = c1 y1 (t) + c2 y2 (t)
satisfies the differential equation and the initial conditions, if and
only if the Wronskian
W = y1 y20 − y10 y2
is not zero at t0 .
Example 38
In a previos example we found that y1 (t) = e −2t and y2 (t) = e −3t
are solutions of the differential equation
y 00 + 5y 0 + 6y = 0
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
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Solutions of Linear Homogeneus Systems. The Wrosnkian
Solutions of Linear Homogeneus Systems. The
Wrosnkian
The Wronskian of these two functions is
−2t
e
e −3t = −e −5t
W =
−2e −2t −3e −3t Since W is nonzero for all values of t, the functions y1 and y2 can
be used to construct solutions of the given differential equation,
together with initial conditions prescribed at any value of t.
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Solutions of Linear Homogeneus Systems
Solutions of Linear Homogeneus Systems. The Wrosnkian
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Theorem
Suppose that y1 and y2 are two solutions of the differential
equation
L[y ] = y 00 (t) + p(t)y 0 (t) + q(t)y (t) = 0
Then the family of solutions
y = c1 y 1(t) + c2 y2 (t)
with arbitrary coefficients c1 and c2 includes every solution of the
above equation if and only if there is a point t0 where the
Wronskian of y1 and y2 is not zero
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Solutions of Linear Homogeneus Systems
Solutions of Linear Homogeneus Systems. The Wrosnkian
Solutions of Linear Homogeneus Systems. The
Wrosnkian
The aboveTheorem states that, if and only if the Wronskian ofy 1
and y2 is not everywhere zero, then the linear combination
c1 y 1(t) + c2 y2 (t)
contains all solutions of the differential equation
y 00 (t) + p(t)y 0 (t) + q(t)y (t) = 0
The solutions y1 and y2 are said to form a fundamental set of
solutions of the above equation if and only if their Wronskian is
nonzero. In this direction we have the following result.
Dr. Marco A Roque Sol
Ordinary Differential Equations