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BC Exam
Differential equations are among the most powerful tools we have for
analyzing the world mathematically. They are used to formulate the
fundamental laws of nature (from Newton’s Laws to Maxwell’s
equations and the laws of quantum mechanics) and to model the most
diverse physical phenomena. This chapter provides an introduction to
some elementary techniques and applications of this important subject.
A differential equation is an equation that involves an unknown
function y and its first or higher derivatives. A solution is a function
y = f (x) satisfying the given equation. As we have seen in previous
chapters, solutions usually depend on one or more arbitrary constants
(denoted A, B, and C in the following examples):
The first step in any study of differential equations is to classify the
equations according to various properties. The most important attributes
of a differential equation are its order and whether or not it is linear.
The order of a differential equation is the order of the highest
derivative appearing in the equation. The general solution of an
equation of order n usually involves n arbitrary constants. For
example,
y " y  0
has order 2 and its general solution has two arbitrary constants A and B.
Just like the order of a Taylor Polynomial!
A differential equation is called linear if it can be written in the form
an  x  y
 n
 an 1  x  y
 n 1
   a1  x  y ' a0  x  y  b  x 
The coefficients aj(x) and b(x) can be arbitrary functions of x, but
a linear equation cannot have terms such as y3, yy ' , or siny.
Separation of Variables
Separable Equations have the form
dy
 f  x g  y .
dx
dy

  sin x  y is separable.
First-Order
dx
dy

 x  y is not separable because x  y is not the product of f  x  & g  y  .
dx
dy
Show that y  x  0 is separable but not linear. Then find the
dx
Here we go ;-)
general solution and plot the family of solutions.
y 2 x2
 ydy   xdx  2  2  C
y 2  x2  C  y   x2  C
yy '  NL
2
2
y
x
 2 1
2
a b
A differential equation is called linear if it can be written in the form
an  x  y  n   an 1  x  y  n 1    a1  x  y ' a0  x  y  b  x 
This is a conic... A family of Hyperbolas
Solve for y
dy
Solutions y  x  C to y  x  0.
dx
2
A Degenerate Hyperbola
Let c  0
y   x2  C
Although it is useful to find general solutions, in applications we are
usually interested in the solution that describes a particular physical
situation. The general solution to a first-order equation generally
depends on one arbitrary constant, so we can pick out a particular
solution y(x) by specifying the value y(x0) for some fixed x0. This
specification is called an initial condition. A differential equation
together with an initial condition is called an initial value problem.
Family of Solutions
to a Particular
Differential Equation
Initial Value Problem Solve the initial value problem
y '  ty,
y 0  3
dy
dy
 ty 
 tdt
dt
y
y  3e
t2

2
dy
t2

   tdt  ln y    C
y
2
 y e
t2
 C
2
 y  Ce
 y  3e
t2

2
t2

C
2
 y e e
y  Ce
t2

2
Family of Solutions
, y  0   3  3  Ce0  C  3
 t 2 /2
Laws of Exponents
y  ec f  t  
y  ec f  t  or  ec f  t 
Since C is arbitrary, eC represents an arbitrary
positive number, and ±eC is an arbitrary
nonzero number. We replace ±eC by C and write
the general solution as
In the context of differential equations, the term “modeling” means finding a differential
equation that describes a given physical situation. As an example, consider water leaking
through a hole at the bottom of a tank. The problem is to find the water level y (t) at time t. We
solve it by showing that y (t) satisfies a differential equation.
Differential Equations  UP
The key observation is that the water lost during the interval from t to t + Δt can be computed in
two ways. Let
v  y   velocity of water flowing through the hole
when the tank is filled to height y
B  area of the hole
Not constant,
but close
A  y   area of horizontal cross-section of the tank
at height y
First, we observe that the water exiting through the hole
during a time interval Δt forms a cylinder of base B and
height υ(y)Δt.
d  rt
V  BV  y  t
Water leaks out of a tank
through a hole of area B at
the bottom.
In the context of differential equations, the term “modeling” means finding a differential
equation that describes a given physical situation. As an example, consider water leaking
through a hole at the bottom of a tank. The problem is to find the water level y (t) at time t. We
solve it by showing that y (t) satisfies a differential equation
The key observation is that the water lost during the interval from t to t + Δt can be computed in
two ways. Let
v  y   velocity of water flowing through the hole
when the tank is filled to height y
B  area of the hole
A  y   area of horizontal cross-section of the tank
at height y
Second, we note that the water level drops by an
amount Δy during the interval Δt.
 water lost between t and t  t  A  y  y
A  y  y  Bv  y  t
y Bv  y 
dy Bv  y 



t
A y
dt
A y
Now we can set up our
differential equation!
Water leaks out of a tank
through a hole of area B at
the bottom.
dy Bv  y 

dt
A y
Velocity of the water passing through the hole is...
To use our differential equation, we need to know
the velocity of the water leaving the hole. This is
given by Torricelli’s Law with (g = 9.8 m/s2):
Given v  y    2 gy  4.43 y m/s
Now we simply plug in our known
values and solve the differential equation
using separation of variables.
dy Bv  y 

dt
A y
v  y    2 gy  4.43 y m/s
Application of Torricelli’s Law
A cylindrical tank of height 4 m and radius 1 m is filled with water.
Water drains through a square hole of side 2 cm in the bottom.
Determine the water level y(t) at time t (seconds). How long does
it take for the tank to go from full to empty?
Solution We can use units of centimeters 1 m  100 cm .
is constant...
A  y    r 2  10, 000 cm 2
2
2
g  9.8 m/s  980 cm/s
B  4 cm 2
 v  y    2  980  y  44.3 y cm/s


4 44.3 y
dy Bv  y 


 0.0056 y
dt
A y
10, 000
dy
dy
 0.0056 y 
 0.0056dt
dt
y

y  0 
dy
   0.0056dt  2 y1/ 2  0.0056t  C
y
y  0.0028t  C  y   C  0.0028t 
2
Step 2. Use the initial condition (the tank was full).
y  0   400 cm  400  C 2  C  20
400
y  t    20  0.0028t 
2
200
400
Which sign is correct?
y  t    20  0.0028t 
2
200
10, 000
Separation of Variables
This is actually an IVP.
10, 000
 te  7143 s
Why can't we just find
dy
 dt ?
CONCEPTUAL INSIGHT The previous example highlights the need
to analyze solutions to differential equations rather than relying on
algebra alone. The algebra seemed to suggest that C = ±20, but further
analysis showed that C = −20 does not yield a solution for t ≥ 0. Note
also that the function
y (t) = (20 − 0.0028t)2
is a solution only for t ≤ te—that is, until the tank is empty. This
function cannot satisfy our original differential equation for t > te
because its derivative is positive for t > te, and solutions of the given
differential must have nonpositive derivatives.
 Only Separation of Variables is tested on the BC Exam.