Lesson Module 4
Differential Equations
I.
Terminology
A.
A differential equation is any equation that contains derivatives.
B.
The order of a differential equation is the order of the highest
derivative in the equation.
Example: The following differential equation is a 3rd order differential
equation because the highest derivative is the third derivative with
respect to time.
d3x
d2x
dx
3t 3  7 2  2  3x  0
dt
dt
dt
C.
A differential equation in which no variable or derivative is higher
than the 1st degree is called a linear differential equation.
Example: The previous differential equation is a linear differential
equation. The following is a non-linear differential equation because
the second derivative with respect to time is raised to the 2 nd degree.
2
 d2x 
 2   3 x  0
 dt 
II.
Linear Differential Equations
Linear differential equations are extremely important in physics and
engineering because there are solution methods that have been
developed to handle various types of linear differential equations. Of
special importance in mechanics is the solution of 2nd order linear
differential equations since Newton’s 2nd Law is a 2nd order
differential equation for position with respect to time.
A.
General Form
Any linear differential equation of order “n” can be written as
dn x
d n 1x
dx
a n t  n  a n 1 (t) n 1  ....... a1 (t)  a 0 (t) x  b(t) .
dt
dt
dt
If b(t) is zero then the differential equation is said to be homogeneous,
otherwise it is inhomogeneous.
B.
Properties of Linear Differential Equations
1.
If two functions f1 and f2 are solutions to a linear differential
equation then g = f1+f2 is also a solution of the linear
differential equation. (Superposition Principle)
Many physical systems obey the superposition principles
including problems involving electric fields, forces, quantum
wave functions, etc. Thus, the differential equations which
describe theses systems must be linear differential equations.
2.
If the function f1 is a solution to a linear differential equation
then the function g = c f1 where C is a non-zero constant is also
a solution of the linear differential equation.
Example: For the following differential equation
d2x
k2 x 0
2
dt
with k2 being a positive constant, we can show by direct substitution
that two solutions to the linear differential equation are
f1  A e kt and f 2  B e -kt
Thus, the most general solution can be written as a sum of the two
independent solutions:
g  A e kt  B e -kt
3.
C.
If xi(t) is the solution to a linear inhomogeneous differential
equation and xh(t) is the solution to the corresponding
homogeneous differential equation then x(t) = xh(t) + xi(t) is
also a solution to the linear inhomogeneous differential
equation.
Solving Inhomogeneous Equations
The total solution to an inhomogeneous differential equation can be
found by the following steps:
1.
Set b(t) = 0 and solve the homogenous differential equation to find
xh(t).
2.
We now solve the inhomogeneous differential equation to find xi(t).
3.
The complete solution is x(t) = xh(t) + xi(t).
The solution contains two parts. The xh(t) part of the solution is
usually known as the transient response of the system. This makes
sense in electrical engineering where this part of the solution usually
fades out due to dissipative elements (resistance, emission antenna,
etc) in the circuit. Personally, I prefer to think of it as the “natural
response” since the response will not fade out if dissipative elements
(resistance, friction, etc.) are either negligible or the loss is
compensated by other energy sources. In classical mechanics, it can
be thought as the way that the system would respond if you hadn’t
applied the driving force b(t). This part of the solution contains all
dependence on any initial conditions of the system. The xi(t) part of
the solution is called the “steady state” solution since it is the part of
the solution that would remain if the transient part faded away after a
long period of time. This part of the solution is totally determined by
the driving force and independent of the initial conditions of the
system.
III.
Constant Coefficient Linear Homogenous Differential Equations
This is an important class of differential equations which we know
how to solve. Physicists often can simplify problems by making
suitable approximations (neglect air drag, etc) until they can obtain
this type of differential equation. The goal of a physicist is to
understand the general behavior of some physical system and to
obtain the laws of physics which govern the system and not to solve a
particular problem to high precision. However, engineers must solve
“real world” problems without excessive simplification (air drag often
can’t be neglected). They often have more difficult differential
equations to solve and must resort to more powerful techniques or
numerical methods than the method in this section to work their
problems.
A.
Solution Principle
The only function whose derivative is the same function times a
constant is an exponential function like the one below where A and k
are constants.
x  A e kt
Doing the Calculus, we see that taking derivatives of this function is
equivalent to multiplying the function by k!!!
dx
k x
dt
d2x 2
k x
dt 2
dn x n
k x
dt n
This gives us two important pieces of information: 1) One solution to
any constant coefficient linear homogenous differential equation is an
exponential like the one above; 2) Using an exponential function as
the trial solution, we can change the problem from a Calculus problem
to an algebra problem involving the solving of a polynomial in k.
B.
Solution Method
1. Substitute x  A e kt and its derivatives into the differential equation.
2. Solve the resulting algebraic equation for values k.
3. The complete solution is the linear superposition of all possible
solutions found in part 2. (Apply superposition principle).
Example: Solve the following differential equation:
3
d2x
dx

2
 4x 0
dt
dt 2
Solution:
1.
3 k2 x  2k x  4x 0
2.
3 k
2

2k  4 x 0
Since this equation holds for all values of t (i.e. even when the particle
is not at the origin, x=0), the expression in the parentheses must
always be zero!
3 k
2

2k  4 x 0
Using the quadratic formula, we have
k
 2  2 2  434 
23
k
1
 11  1
11


i
3
3
3
3
Thus, we have two unique solutions as expected for this 2nd order
differential equation:
f1
1
t
(t)  A e 3
f2
1
t
(t)  B e 3
i
11
t
3
-i
11
t
3
e
e
Thus, the most general solution to the problem is the sum of our two
possible solutions:
x(t) 
1
t
e3
 i
Ae


11
t
3
 Be
-i
11
t
3




The two arbitrary constants (A and B) depend on the initial conditions
of the system. Since the equation was 2nd order, its solution is
equivalent to having performed two integrations so a general solution
should have two arbitrary constants (integration constants) which
depend on initial conditions.
IV.
Important 2nd Order Differential Equation To Know By Heart
Since Newton’s 2nd Law is a 2nd order differential equation of position
with respect to time, it should come as no surprise that the solution of
2nd order differential equations plays a paramount role in the academic
success of physics and engineering students. Thus, the student should
know the solutions to the most important (by the frequency that they
appear in course work) cases in the same way that they know their
multiplication tables.
A.
The Important Equation
d2x 2
 k x where k2 is a constant.
2
dt
B.
Linear Case : k2 = 0
The solution to this problem is a straight line!! For physics, we know
that this corresponds to a particle with no acceleration so
x  vo t  x o
where I have already interpreted the physical significance of the slope
(vo) and the y-intercept (xo).