2.

3 Linear Equations

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Linear Equations (1 of 19)
A Definition

Definition 2.3.1 Linear Equation

A first-order differential equation of the form

is said to be a linear equation in the variable y.

2
Linear Equations (2 of 19)
Standard Form
By dividing both sides of (1) by the lead coefficient a1(x), we
obtain a more useful form, the standard form, of a linear
equation:

We seek a solution of (2) on an interval I for which both

coefficient functions P and f are continuous.

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Linear Equations (3 of 19)
Before we examine a general procedure for solving
equations of form (2) we note that in some instances (2) can
be solved by separation of variables. For example, you
should verify that the equations

are both linear and separable, but that the linear equation

is not separable.

4
Linear Equations (4 of 19)
Method of Solution
The method for solving (2) hinges on a remarkable fact that
the left-hand side of the equation can be recast into the form
of the exact derivative of a product by multiplying the both
sides of (2) by a special function μ(x). It is relatively easy to
find the function μ(x) because we want

5
Linear Equations (5 of 19)
The equality is true provided that

The last equation can be solved by separation of variables.

Integrating

6
Linear Equations (6 of 19)
Hence we can simplify life and choose c2 = 1. The function

is called an integrating factor for equation (2). We multiplied

both sides of (2) by (3) and, by construction, the left-hand side
is the derivative of a product of the integrating factor and y:

7
Linear Equations (7 of 19)

We can integrate both sides of the last equation,

and solve for y. The result is a one-parameter family of

solutions of (2):

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Linear Equations (8 of 19)
Solving a Linear First-Order Equation
(i) Remember to put a linear equation into the standard form
(2).
(ii) From the standard form of the equation identify P(x) and
then find the integrating factor . No constant need
be used in evaluating the indefinite integral ∫P(x)dx.

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Linear Equations (9 of 19)
(iii) Multiply the both sides of the standard form equation by
the integrating factor. The left-hand side of the resulting
equation is automatically the derivative of the product of
the integrating factor and y:

(iv) Integrate both sides of the last equation and solve for y.

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Example 1 – Solving a Linear Equation
Solve

Solution:
This linear equation can be solved by separation of variables.
Alternatively, since the differential equation is already in
standard form (2), we identify P(x) = −3, and so the
integrating factor is

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Example 1 – Solution (1 of 1)
We then multiply the given equation by this factor and
recognize that

Integration of the last equation,

then yields

12
Linear Equations (10 of 19)
General Solution
Suppose again that the functions P and f in (2) are
continuous on a common interval I. In the steps leading to
(4) we showed that if (2) has a solution on I, then it must be
of the form given in (4).

Conversely, it is a straightforward exercise in differentiation

to verify that any function of the form given in (4) is a
solution of the differential equation (2) on I.

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Linear Equations (11 of 19)
In other words, (4) is a one-parameter family of solutions of
equation (2) and every solution of (2) defined on I is a
member of this family. Therefore we call (4) the general
solution of the differential equation on the interval I.

Now by writing (2) in the normal form y′ = F(x, y), we can

identify F(x, y) = −P(x)y + f(x) and ∂F ∕ ∂y = −P(x).

From the continuity of P and f on the interval I we see that F

and ∂F ∕ ∂y are also continuous on I.

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Linear Equations (12 of 19)
We know that there exists one and only one solution of the
first-order initial-value problem

defined on some interval I0 containing x0.

But when x0 is in I, finding a solution of (6) is just a matter of

finding an appropriate value of c in (4)—that is, to each x0 in
I there corresponds a distinct c.

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Example 3 – General Solution
Solve

Solution:
Dividing by x, the standard form of the given DE is

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Example 3 – Solution (1 of 2)
From this form we identify P(x) = −4 ∕ x and and
further observe that P and f are continuous on

Hence the integrating factor is

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Example 3 – Solution (2 of 2)
Here we have used the basic identity Now
we multiply (7) by and rewrite

It follows from integration by parts that the general solution

defined on the interval

18
Linear Equations (13 of 19)
Except in the case in which the lead coefficient is 1, the
recasting of equation (1) into the standard form (2) requires
division by a1(x). Values of x for which a1(x) = 0 are called
singular points of the equation.

Singular points are potentially troublesome. Specifically, in

(2), if P(x) (formed by dividing a0(x) by a1(x)) is
discontinuous at a point, the discontinuity may carry over to
solutions of the differential equation.

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Example 5 – An Initial-Value Problem
Solve

Solution:
The equation is in standard form, and P(x) = 1 and f(x) = x
are continuous on

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Example 5 – Solution (1 of 2)
The integrating factor is so integrating

gives

Solving this last equation for y yields the general solution

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Example 5 – Solution (2 of 2)
But from the initial condition we know that y = 4 when x = 0.
Substituting these values into the general solution implies
that c = 5. Hence the solution of the problem on the interval

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Linear Equations (14 of 19)
Figure, obtained with the aid of a graphing utility, shows the
graph of the solution (9) in dark blue along with the graphs
of other members of the one-parameter family of solutions

Figure 2.3.2

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Linear Equations (15 of 19)
It is interesting to observe that as x increases, the graphs of
all members of this family are close to the graph of the
solution y = x − 1.

The last solution corresponds to c = 0 in the family and is

shown in dark green in Figure 2.3.2. This asymptotic
behavior of solutions is due to the fact that the contribution
of , c ≠ 0, becomes negligible for increasing values of x.

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Linear Equations (16 of 19)
We say that is a transient term, since
While this behavior is not characteristic of all general
solutions of linear equations, the notion of a transient is
often important in applied problems.

Piecewise-Linear Differential Equation

In the construction of mathematical models (especially in
the biological sciences and engineering) it can happen that
one or more coefficients in a differential equation is a
piecewise-defined function.

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Linear Equations (17 of 19)
In particular, when either P(x) or f(x) in (2) is a piecewise-
defined function the equation is then referred to as a
piecewise-linear differential equation.

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Example 6 – An Initial-Value Problem
Solve

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Example 6 – Solution (1 of 4)
The graph of the discontinuous function f is shown in figure.

Figure 2.3.3

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Example 6 – Solution (2 of 4)
We solve the DE for y(x) first on the interval [0, 1] and then
on the interval For 0 ≤ x ≤ 1 we have

Integrating this last equation and solving for y gives

Since y(0) = 0, we must have c1 = −1, and
therefore 0 ≤ x ≤ 1.

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Example 6 – Solution (3 of 4)
Then for x > 1 the equation

leads to Hence we can write

By appealing to the definition of continuity at a point, it is

possible to determine c2 so that the foregoing function is
continuous at x = 1.

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Example 6 – Solution (4 of 4)
The requirement that implies that
or c2 = e − 1. As seen in figure, the
function

Graph of (10) in Example 6

Figure 2.3.4

is continuous on

Note: Can we call (10) a solution of the given IVP? Why?

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Linear Equations (18 of 19)
Error Function
In mathematics, science, and engineering some important
functions are defined in terms of nonelementary integrals.
Two such special functions are the error function and
complementary error function:

From the known result we can write

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Linear Equations (19 of 19)
Using the additive interval property of definite integrals
we can rewrite the last result in the alternative
form

It is seen from (12) that the error function erf(x) and

complementary error function erfc(x) are related by the
identity

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Example 7 – The Error Function
Solve the initial-value problem

Solution:
The differential equation is already in standard form, and so
we see that the integrating factor is

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Example 7 – Solution (1 of 4)
Multiplying both sides of the equation by this factor then
gives which is the same as

Because indefinite integration of both sides of equation (13)

leads to the nonelementary integral dx we identify x0 = 0
and use definite integration over the interval [0, x]:

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Example 7 – Solution (2 of 4)
Using the initial condition y(0) = 1 the last expression yields
the solution

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Example 7 – Solution (3 of 4)
Then by inserting the factor into this solution in the
following manner—

we see from (11) that (14) can be rewritten in terms of the

error function as

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Example 7 – Solution (4 of 4)
The graph of solution (15), obtained with the aid of a CAS,
is given in figure.

Figure 2.3.5

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