Suppose that you are driving along a road and you want to predict where you will be in the future. You need to know two things--
Example:
A cup of coffee in a room is cooling according to the differential equation
dT
-- = 0.15 (68 - T)
dt
Its temperature at time t = 0 is 175 degrees Fahrenheit. Predict its future temperature.
The function
-0.15 t
T(t) = 68 + C e
where C is any constant, is the solution of this differential equation (verify this fact). We use the initial condition T(0) = 175 to determine the value of the constant C as follows.
-0.15 (0)
T(0) = 68 + C e
175 = 68 + C
C = 107
So the function is
-0.15 t
T(t) = 68 + 107 e
Example:
An object is dropped from a height of six feet. Its acceleration is described by the differential equation
Describe the height y(t) of the object t seconds after it is dropped.
Since y'' = -32 we see that
where a is any constant. This implies that
where b is another constant. Notice that we have two constants because the differential equation was a second order differential equation.
Since the ball was dropped at time t = 0 from a height of six feet we know that
The second initial condition, y'(0) = 0, comes from the fact that the ball was dropped--that is, released, at time t = 0. If it had been thrown upward with a velocity of 10 feet per second then this initial condition would have been y'(0) = 10.
Now we use these initial conditions to find the values of the two constants as follows.
y (0) = 6
2
-16 (0) + a (0) + b = 6
b = 6
y'(t) = -32 t + a
y'(0) = -32 (0) + a = 0
a = 0
So we have
2
y(t) = -16 t + 6.
The preceding examples are typical. The solution of a first order differential equation has one constant whose value is determined by one initial condtion, the solution of a second order differential equation has two constants whose values are determined by two initial conditions, and in general an n-th order differential equation has n constants whose values are determined by n initial conditions.
A solution of a differential equation with its constants undetermined is called a general solution. The solution of an IVP complete with the values of the constants is called a particular solution. For example, the general solution of the differential equation
is
and the particular solution of the IVP
is
In practical problems there may be additional constants besides the ones that come from the differential equation. Consider the following example.
Example:
An object is cooling in a room and its temperature change may be described by Newton's Model of Cooling. The temperature of the room is constant. The following table gives the temperature of the object at three times
-----------------
Time Temperature
-----------------
0 80.00
5 70.00
10 62.00
-----------------
Since the temperature change can be described by Newton's Model of Cooling we know that it obeys the differential equation
dT
-- = k(A - T)
dt
where k is a constant. We are also given that the room temperature, A, is constant. Thus, we have two constants in addition to the constant that comes from the fact that this is a first order differential equation. The solution of this equation is
We can use the three data points in the table to determine the three constants as follows.
-0k
T(0) = A + C e
80.00 = A + C (1)
-5k
T(5) = A + C e
-5k
70.00 = A + C e (2)
-10k
T(10) = A + Ce
-10k
62.00 = A + Ce (3)
Subtracting equation (2) from equation (1) we obtain
-5k
10 = C - C e (4)
Subtracting equation (3) from equation (1) we obtain
-10k
18 = C - Ce (5)
Now multiplying equation (4) by 1.8 we obtain
-5k
18 = 1.8 C - 1.8 Ce (6)
and we see that
-10k -5k
C - Ce = 1.8 C - 1.8 Ce
-10k -5k
1 - e = 1.8 - 1.8e
-10k -5k
e - 1.8e + 0.8 = 0 (7)
Now if we let
-5k
x = e then
2 -10k
x = e
and equation (7) becomes
2
x - 1.8 x + 0.8 = 0
Now, by the quadratic formula
x = 1 or x = 0.8 and thus,
-5k
e = 0.8
-5k = ln 0.8
-5k = -0.223144
k = 0.044629
or
-5k
e = 1
-5k = ln 1
-5k = 0
k = 0
It is physically impossible for k to be zero, so the solution we seek is k = 0.044629.
Now that we know k it is easy to determine A and C. Equation (1) is
A + C = 80.00 (8)
and equation (2) is
-5k
A + Ce = 70
A + 0.8 C = 70 (9)
Subtracting equation (9) from equation (8) we obtain
0.2 C = 10
C = 50
and by equation (8)
A + 50 = 80
A = 30
So, finally, we have
-0.044629t
T(t) = 30 + 50 e
For each of the following problems check that the indicated function is a solution of the differential equation and then find the values of the constants needed to solve the initial value problem.
y'' + 4y = 0, y(0) = 1, y'(0) = 0 y = A sin 2t + B cos 2t
y'' + 4y = 0, y(0) = 0, y'(0) = 1 y = A sin 2t + B cos 2t
y'' + 4y = 0, y(0) = 2, y'(0) = 3 y = A sin 2t + B cos 2t
y' = ky, y(0) = 1, y'(0) = 1
kt
y = C e
y' = ky, y(0) = 2, y'(0) = 5
kt
y = C e
y' = ky, y(0) = 2, y(1) = 5
kt
y = C e
that describes the height of a falling object. Then find the particular solution that satsifies the conditions below.
that describes the height of a falling object. Then find the particular solution that satsifies the conditions below.
Consider the differential equation
dT
-- = k(70 - T)
dt
where k is a positive constant. The solution of this differential equation is
-kt
T = 70 + C e
This differential equation describes the heating or cooling of an object in a room whose temperature is a constant 70 degrees. This is the reason that the constant k must be positive. Since the function T(t) has two constants we expect to be able to satisfy two conditions. The purpose of this set of problems is to investigate this statement -- "we expect to be able to satisfy two conditions."
dT -- = k(70 - T), T(0) = 100, T'(0) = -5 dt
dT -- = k(70 - T), T(0) = 100, T'(0) = 5 dt
dT -- = k(70 - T), T(0) = 50, T'(0) = 5 dt
dT -- = k(70 - T), T(0) = 50, T'(0) = -5 dt
dT -- = k(70 - T), T(0) = a, T'(0) = b dt
Notice that your answers to the five questions above depend on whether you require the constant k to be positive.
dT -- = k(70 - T), T(0) = 100, T(50) = 80 dt
dT -- = k(70 - T), T(0) = 100, T(50) = 60 dt
dT -- = k(70 - T), T(0) = 40, T(50) = 60 dt
dT -- = k(70 - T), T(0) = 40, T(50) = 80 dt
dT -- = k(70 - T), T(0) = a, T(50) = b dt
Explain your answer physically.