Calculus 1 Problems & Solutions Chapter 3 Section 3.9 |
|
3.9
|
Return To
Contents
Go To Problems & Solutions
|
Review |
|
1. Differential Equations |
Let y = x2. Then y' = 2x. The
equation y' = 2x involves
the derivative of a function, and for this reason is called a
differential equation. Now, to work in reverse, suppose a function y of x is unknown
but its derivative is known to be y'
=
2x.
This is a differential equation. Clearly an antiderivative of 2x is x2. If y = x2,
then y' = 2x. The
function y
= x2 is a
function that satisfies or solves the differential equation y' = 2x, and for
this reason is called a solution of
that differential equation. The equation y'' = x 1 is also a differential
equation. Note that the equations y'
= 2x
and y'' = x 1 can
also be written as y' 2x = 0 and y'' x + 1 = 0
respectively.
In Section
3.8 Example 5.1, we were given x''
= 4, and in Part
6 of the same section, we had y''
= g.
The equations
x'' = 4 and y'' = g are
examples of differential equations.
Now suppose we have a function y that's
unknown and that has derivatives y'
and y'' that are also
unknown, but all three
satisfy the equation x2y'' xy' 3y = 0. This is a differential
equation too. Any function that satisfies or solves it is called a solution
of it.
The function y = x2 is a solution of the differential equation y' = 2x, and so are the functions y = x2 + 1, y = x2 200, or y = x2 + C for any constant C, because the derivative of a constant is 0. Remember that a function y = f(x) can be considered as its own 0th (zeroth) derivative.
|
An equation that involves derivatives of one or more
orders of an unknown function is called a differential equation.
|
For the derivatives in a
differential equation, notations other than the prime (', '', etc) notations are also used. For
example, the following three equations are the same equation:
x2y'' xy' 3y = 0,
In this section we investigate two
types of differential equations: y'
= f(x) and y'' = f(x), where f is a known
function
(not the unknown function y). For the
remainder of this section, the abbreviation DE stands for differential
equation.
Go To Problems & Solutions Return To Top Of Page
Suppose the function y = F(x) is a
solution of the DE y' = f(x), ie F'(x) = f(x). Then so
are y
= F(x) + 2 and y =
F(x) 100/7,
as the derivative of a constant is 0. In fact, for any constant C, the
function y
= F(x) + C is a
solution. We
see that there are infinitely many solutions. Any two solutions F(x) + C1 and F(x) + C2
of the DE y' = f(x) differ by
a
constant, which here is C1 C2.
We use the function y
= F(x) + C, where C is an
arbitrary constant, to represent the
collection of all the solutions, and call it the general solution of the
DE y' = f(x). To solve
a differential equation means
to find its general solution.
Solving y' = f(x)
If y' = f(x) then f is the derivative of y so y is an
antiderivative of f.
Thus solving the DE y' = f(x) clearly
amounts to
finding the general antiderivative of f(x). See Section
3.7 Definition 3.1. If f
has an antiderivative, then the DE y'
= f(x)
has that antiderivative as a solution of it and the general antiderivative of f as its
general solution.
Example 2.1
Solve y' = x3 5x2 + 2x 4.
Solution
EOS
Graphs Of Solutions
Any two solutions F(x) + C1 and F(x) + C2 of the DE y' = f(x) differ by a constant, which here is C1 C2. So, geometrically, the set of the graphs of F(x) + C obtained by running the constant C thru the set R of real numbers can be visualized as a collection of graphs congruent to each other. Some members of this collection are sketched in Fig. 2.1. Also see Section 3.7 Graphs Of Antiderivatives. Let x1 be any point of dom( f ) and F1 and F2 any 2 solutions. We have F1'(x1) = f(x1) = F2'(x1). This shows that the graphs of F1 and F2 have the same slope, ie, have parallel tangent lines, at x1. At any point x the graphs of all the solutions of the DE y' = f(x) have the same slope or parallel tangent lines.
|
|
Graphs Of Some Solutions Of y' = f(x). |
It's worth repeating that any 2
solutions of y' = f(x) differ by
a constant. If F(x) is a
solution then any other solution can
be obtained by adding a constant to F(x).
Go To Problems & Solutions Return To Top Of Page
|
3. The Differential Equation y'' = f(x) |
Example 3.1
Solve the differential equation y'' = 5x.
Solution
EOS
a.C1 and C2
are arbitrary constants. The general solution of y'' = 5x is y = (5/6)x3 + C1x + C2.
To verify the answer y
= (5/6)x3 + C1x + C2 we just differentiate it, twice: y' = (5/2)x2 + C1,
y'' = 5x. The
general solution of the DE y'' =
f(x) is of the
form y
= G(x) + C1x + C2,
where G''(x) = f(x) and C1 and C2
are arbitrary constants.
b. We saw above that any 2
solutions of the DE y' = f(x) differ by
a constant. Now let's check to see if the same is true
for the solutions of the DE y'' = f(x). Let F1(x)
= G(x) + C1x + C2
and F2(x) = G(x) + D1x + D2 be any 2 of its
solutions. Clearly they differ by a constant if C1 = D1.
Now let's check:
F1(x) F2(x) = (G(x) + C1x + C2) (G(x) + D1x + D2) = (C1 D1)x + (C2 D2).
Thus indeed
they differ by a constant if C1 D1
= 0 or C1 = D1,
ie if they are of the form F1(x) = G(x) + Cx + K1
and F2(x) = G(x) + Cx + K2;
otherwise they don't.
We see that any
2 solutions of y'' = f(x) differ by
a linear function including a constant (a constant function is a linear
function). If F(x) is a solution of y'' = f(x) then any other solution can be
obtained by adding a linear function including
a constant to F(x).
Go To Problems & Solutions Return To Top Of Page
|
4. Initial-Value Problems |
Note 4.1
In Section
3.8 Example 5.1, we solved the DE x'' = 4 subject to two initial conditions x'(0) = 0 and x(0) = 0, and
we got
x =
2t2. In Part
6 of the same section, we solved the DE y'' = g subject to two initial conditions y'(0) (= v(0)) = v0 and
y(0)
= y0, and we got y = (1/2)gt2 + v0t + y0. The problem of solving a DE subject to
one or more prescribed values
for the solution function or its derivatives is referred to as an initial-value
problem.
The Case Of y' = f(x)
As we saw above, the DE y' = f(x) has
infinitely many solutions, the graphs of some of which are illustrated in Fig. 2.1.
However, as seen there, given a point (x0,
y0) [y0 is not f(x0)],
there's only one graph that passes thru the point (x0, y0).
If y
= F1(x) is the solution whose graph is
that one that passes thru the point (x0,
y0), then we must have y0 = F1(x0).
Now suppose we're given the DE y' = f(x) and asked
to find the particular solution whose graph passes thru a given point
(x0, y0),
ie, the solution y
= y(x) such that y(x0) = y0.
This type of problem is called an initial-value problem, because
the point (x0, y0)
is considered as the initial point of the solution. For the reason for the use
of the adjective initial, see
Note 4.1 above and Remarks
4.2 iii below.
Example 4.1
Solve the initial-value problem:
Solution
EOS
Remark 4.1
For clarity, an initial-value problem is stated in the following format:
Solve the initial-value problem:
To solve this problem, first we
find the general solution y
= F(x) + C where C is an
arbitrary constant. Next, knowing
that the wanted particular solution has equation y = F(x) + C for some
constant C
and has value y
= y0 at x = x0,
we
substitute x
= x0 and y = y0
in this equation to get y0 = F(x0)
+ C,
which we solve for C
to obtain C
= y0 F(x0).
Then
we substitute this value of C
in the equation y
= F(x) + C to arrive
at y
= F(x) + y0 F(x0),
which is the wanted solution.
The Case Of y'' = f(x)
Again consider the DE y''= 5x. As seen in
Example 3.1, its general solution is y = (5/6)x3 + C1x + C2, where there are two
arbitrary constants, C1 and C2.
One way to obtain a particular solution is to specify two initial conditions to
determine the
two constants, as shown in the following example.
Example 4.2
Solve the initial-value problem:
Solution
EOS
Remarks 4.2
a. The initial conditions y'(1) = 2 and y(1) = 1 describe the behavior of the solution at the same point, x = 1, which is considered as the initial point of the solution.
b. First, from the equation y'' = 5x we get y' = (5/2)x2 + C1, which represents a collection of its infinitely many antiderivatives whose graphs are congruent to each other. The condition y'(1) = 2 tells us to select the particular antiderivative whose graph passes thru the point (1, 2). That antiderivative is y' = (5/2)x2 1/2.
Next, from that antiderivative y' = (5/2)x2 1/2 we get its own general antiderivative y = (5/6)x3 (1/2)x + C2, which represents a collection of its infinitely many antiderivatives whose graphs are congruent to each other. The
condition y(1) = 1 tells us to select the particular antiderivative whose graph passes thru the point (1, 1). That antiderivative is y = (5/6)x3 (1/2)x 4/3. It's the desired solution.
c. The adjective initial undoubtedly first appeared in the study of
motion where it was used to describe initial conditions, ie conditions at
initial time t0 = 0. See Section
3.8 Remarks 5.1 iii and Remarks
6.1 iv of the same section. However, presently the phrase initial
conditions is utilized to refer
to conditions of the unknown function or its derivatives at any
single value of the independent variable.
Go To Problems & Solutions Return To Top Of Page
|
5. Boundary-Value Problems |
Since the general solution of
the DE y'' = f(x) is of the
form y
= G(x) + C1x + C2
(see Remarks 3.1), another way to obtain a particular
solution is to generate from this equation a system of two equations in the
unknowns C1
and C2
that
Example 5.1
Solve the boundary-value problem:
Multiplying the first equation by
1 and adding the resulting equation to the second we obtain 2C1 = 82,
so that C1 =
41 . Then C2= 1 C1
= 1 (41) = 42. Thus the solution is y = x4
41x
+ 42.
EOS
Remark 5.1
The boundary conditions y(1) = 2 and y(3) = 0
describe the behavior of the solution at different points, x1
= 1 and x2 =
3, which are viewed as the boundary points of the solution.
|
Problems & Solutions |
1. Solve the following differential equations.
Solution
2. Solve the following differential equations.
a. y'' = 0.
Solution
3. Solve the following initial-value problems.
Solution
4. Solve the following initial-value problems.
Solution
5. Suppose the
differential equation y'' =f(x) has a
solution y
= F(x) on [a, b], where a < b. Show that
for any real
Solution
Return To Top Of Page Return To Contents