Return To Contents
Go To Problems
& Solutions
1. Differential Equations 
In Section
5.8 Example 5.1, we were given x'' =
4, and in Part
6 of the same section, we had y'' =
g. We solved these
equations to determine the functions x and y
respectively. These are situations where a quantity isn't known but its second
derivative is known, and our task is to determine the quantity. There are
situations where a quantity isn't known but its rate of
change (first derivative) is known or given, and our task is to determine the
quantity. For example, we may have to determine
y if dy/dx = cos x. There are also situations where a quantity
isn't known but it and its derivatives of order 1 or higher are
related by an equation, and our task is to determine the quantity. For example,
we may have to determine y if y'' + 3y' 2y =
0. Recall that a function is considered as its own zeroth (0th) derivative: y^{(0)} = y. An equation involving derivatives of order
1 or
higher of a function is called a differential equation. To solve a differential
equation is to determine the function that solves or
satisfies the equation (it and its derivatives solve or satisfy the equation).
Let y = x^{2}. 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 x^{2}. If y = x^{2}, then y' =
2x. The function y = x^{2} 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.
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 x^{2}y'' 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 = x^{2} is a solution of the differential equation y' = 2x, and so are the functions y = x^{2} + 1,
y = x^{2} 200, or y =
x^{2} + C for any constant C,
because the derivative of a constant is 0. Any solution can be obtained from
the solution y = x^{2} + C
by giving a particular value to C. For this reason, y = x^{2} + C is called the general solution. It
represents all the solutions. A
solution with a particular value of C is called, well, a particular solution.
Remember that a function y = f(x) can be considered
as its own 0th (zeroth) derivative.
Consider the differential equation y'' = 2x. Then y' = x^{2} + C_{1} and y = (x^{3}/3) + C_{1}x + C_{2}. The function y = (x^{3}/3) + C_{1}x + C_{2},
where C_{1} and C_{2} are arbitrary constants,
is the general solution of the differential equation y'' = 2x.
An equation that involves derivative(s) of order(s) 1 and/or higher
of an unknown function is called a differential equation. 
For the derivatives in a differential
equation, notations other than the prime notations ( ', '', etc ) are also used. For example,
the following three equations are the same equation:
x^{2}y'' 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.
Equations that involve only derivatives
of functions of one variable are called ordinary differential equations.
Equations that
involve partial derivatives of functions of several (two or more) variables are
called partial differential equations. In this
chapter we of course touch on the surface of only ordinary DEs.
Go To Problems & Solutions Return To Top Of Page
2. The Differential Equation y' = f (x) 
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) + C_{1} and F(x) + C_{2} of the DE y' = f(x) differ by a
constant, which here is C_{1} C_{2}. 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 5.7 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' = x^{3} 5x^{2} + 2x 4.
Solution
EOS
Graphs Of Solutions
Any two solutions F(x) + C_{1} and F(x) + C_{2} of the DE y' = f(x) differ by a constant, which here is C_{1} C_{2}. 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 5.7 Graphs Of Antiderivatives. Let x_{1} be any point of dom( f ) and F_{1} and F_{2} any 2 solutions. We have F_{1}'(x_{1}) = f(x_{1}) = F_{2}'(x_{1}). This shows that the graphs of F_{1} and F_{2} have the same slope, ie, have parallel tangent lines, at x_{1}. 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. C_{1} and C_{2} are arbitrary
constants. The general solution of y'' =
5x is y = (5/6)x^{3} + C_{1}x + C_{2}. To verify the answer y
= (5/6)x^{3} + C_{1}x + C_{2} we just differentiate it, twice: y' = (5/2)x^{2} + C_{1}, y'' =
5x. The general solution of the DE y'' =
f(x) is of the form y = G(x) + C_{1}x + C_{2}, where G''(x) = f(x) and C_{1} and C_{2} 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 F_{1}(x) = G(x) + C_{1}x + C_{2} and F_{2}(x) = G(x) + D_{1}x + D_{2} be any 2 of its
solutions. Clearly they differ by a
constant if C_{1} = D_{1}. Now let's check:
F_{1}(x) F_{2}(x) = (G(x) + C_{1}x + C_{2}) (G(x) + D_{1}x + D_{2}) = (C_{1} D_{1})x + (C_{2} D_{2}).
Thus indeed they differ by a
constant if C_{1} D_{1} = 0 or C_{1} = D_{1}, ie if they are of the form F_{1}(x) = G(x) + Cx + K_{1}
and F_{2}(x) = G(x) + Cx + K_{2}; 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. InitialValue Problems 
Note 4.1
In Section
5.8 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 = 2t^{2}. In Part
6 of the same section, we solved the DE y'' = g subject to two initial conditions y'(0) (= v(0)) = v_{0} and
y(0) = y_{0}, and we got y = (1/2)gt^{2} + v_{0}t + y_{0}. 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 initialvalue
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 (x_{0}, y_{0}) [ y_{0} is not f(x_{0})], there's only one graph that passes thru
the point (x_{0}, y_{0}).
If y = F_{1}(x) is the solution whose graph is that one
that passes thru the point (x_{0}, y_{0}), then we must have y_{0} = F_{1}(x_{0}).
Now suppose we're given the DE y' = f(x) and asked to find the particular solution
whose graph passes thru a given point
(x_{0}, y_{0}), ie, the solution y = y(x) such that y(x_{0}) = y_{0}. This type of problem is called an initialvalue
problem, because
the point (x_{0}, y_{0}) 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 c below.
Example 4.1
Solve the initialvalue problem:
EOS
a. For clarity, an initialvalue problem is stated in the following format:
Solve the initialvalue problem:
b. From the equation
y' = x 2 we get y = x^{2}/2 2x + C, which represents a collection of infinitely many antiderivatives y of
y';
the graphs of y are congruent to each other. The condition y(1) = 3 tells us to select the particular antiderivative y
whose graph passes thru the point
(1, 3). That antiderivative is y = x^{2}/2 2x + 9/2.
c. In the equation y' = f(x), the highest derivative order is 1, and we need 1 initial condition to solve the initialvalue problem.
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)x^{3} + C_{1}x + C_{2}, where there are two
arbitrary constants, C_{1} and C_{2}. 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 initialvalue problem:
EOS
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)x^{2} + C_{1}, which represents a collection of infinitely
many antiderivatives y'
of y'';
the graphs of y'
are congruent to each other. The condition y'(1) = 2 tells us to select the particular antiderivative
y'
whose graph passes thru the point (1, 2). That antiderivative is y' = (5/2)x^{2} 1/2.
Next, from that antiderivative y' = (5/2)x^{2} 1/2 we get its own general antiderivative y = (5/6)x^{3} (1/2)x + C_{2}, which
represents a collection of infinitely many antiderivatives y of y';
the graphs of y are congruent to each other. The condition
y(1) = 1 tells us to select the particular
antiderivative y whose graph passes thru the point (1, 1).
That antiderivative is y
= (5/6)x^{3} (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 t_{0} = 0. See Section
5.8 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.
d. In the equation y'' = f(x), the highest derivative order is 2, and we need 2 initial conditions to solve the initialvalue problem.
Go To Problems & Solutions Return To Top Of Page
5. BoundaryValue Problems 
Since the general solution of the DE y'' = f(x) is of the form y = G(x) + C_{1}x + C_{2} (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 C_{1} and C_{2} that
Example 5.1
Solve the boundaryvalue problem:
Multiplying the first equation by 1 and adding the resulting equation
to the second we obtain 2C_{1} = 82, so that C_{1} =
41 . Then C_{2} = 1 C_{1} = 1 ( 41) = 42. Thus the solution is y = x^{4} 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, x_{1} = 1 and x_{2} =
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 initialvalue problems.
Solution
4. Solve the following initialvalue 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