## Calculus Of One Real Variable By Pheng Kim Ving Chapter 16: Differential Equations Group 16.1: First-Order Equations Section 16.1.3: First-Order Linear Equations

16.1.3
First-Order Linear Equations

 1. First-Order Linear Differential Equations

In this section, the abbreviation “DE” stands for “differential equation”.

In Section 7.5 Part 3, we encountered the DE dy/dt = ky, or y' + by = 0, where b = – k. In this equation, the highest order of
derivative is 1 (the equation is of first order), and each term contains at most one factor
y or y', but not both, whose power is 1
(the equation is linear). It's called a first-order linear differential equation. Similarly, the equation
x2y - 2x3y' - ex + 3 = 0 is a
first-order linear DE. It can be converted to the form
y' - (1/2x)y = (3 - ex)/2x3 by changing the order of the first 2 terms,
moving the last 2 terms to the right-hand side, and dividing the resulting equation by -2
x3. This form is called the standard
form
of the DE.

##### Definition 1.1 – First-Order Linear Differential Equations
 A DE that can be written in the form:   y' + p(x)y = q(x),   where p(x) and q(x) are given continuous functions, is called a first-order linear differential equation. The form y' + p(x) y = q(x) is called the standard form of the DE.

The highest derivative order is 1, thus the adjective “first-order”, and each term contains at most either a factor y or a factor y'
raised to just the power of 1, thus the adjective “linear” (note: a term such as
y'y is of power 1 + 1 = 2). Remark that the
equation is written in the decreasing order of the derivative (recall that
y = y(0)). The coefficient of y' is 1 and that of y is p(x).
Observe that while the equation is linear, each function
y, y', p(x), and q(x) doesn't have to be linear. For example, y = 2x + 3
is linear while
y = x2 + 3 and q(x) = sin x - 1 aren't.

The DE y' = f (x), as discussed in Section 16.1.1 Part 2, is a first-order linear one. For this equation, p(x) = 0 and q(x) = f (x).

##### When The Coefficient Of y' Isn't 1

Suppose the coefficient of y' of the DE:

f(x)y' + p1(x)y = q1(x)

is the function f(x) that's not the constant function 1. In this case, to get an equation of the form y' + p(x)y = q(x) that has 1
as the coefficient of
y' and that's equivalent to the given equation, we divide both sides of the given equation by f(x):

##### A Particular Case

Consider the simple example of the particular case of the above DE, y' + by = 0, where p(x) = b and q(x) = 0, and where b is
a constant. We solve it as follows:

y' + by = 0

y' = – by,

So y = Cebx + K isn't the general solution of y' + by = 0. The reason is that the general solution of a first-order equation
involves only one arbitrary constant, and
y = Cebx already has one arbitrary constant, which is C. The arbitrary constant C
originates from the arbitrary constant
C1, which is added to -bx to obtain the general antiderivative of -b. This confirms that

Now let's solve the DE by using an alternative approach. The left-hand side of y' + by = 0 has 2 terms, one term contains the
derivative
y' as a factor and the other term contains the function y as a factor. This reminds us of the product rule of
differentiation: (
uv)' = u'v + uv', or (uv)' = uv' + u'v, where the first term of the right-hand side contains the derivative v'
as a factor and the other term contains the function
v as a factor. So we try to examine if y' + by is the derivative of the
product of some function, say
u, and y, so that the product is uy, whose derivative is (uy)' = uy' + u'y. We see that y' + by =
1
y' + by, so it should be that u = 1 and u' = b. But b isn't the derivative of 1 unless b = 0. Thus y' + by, when b is non-0, in
itself can't be the derivative of the product of any function and
y. The problem is that the coefficient of y' is 1. So we multiply
both sides of
y' + by = 0 by a function u, obtaining uy' + buy = 0. In order for uy' + buy to be the derivative uy' + u'y of uy,
bu must be the derivative u' of u. We determine u so that this requirement is fulfilled, as follows:

u' = bu,

Multiplying y' + by = 0 by ebx works out well. The general solution is the same as obtained previously.

##### General Case

We don't need to memorize the formula of the general solution y of the DE y' + p(x)y = q(x). In solving such an equation,
we instead can follow the procedure that leads to that formula.

##### Integrating Factors

In the general solution:

The solution procedure of the first-order linear DE can be summarized as follows.

Solution Procedure Of The First-Order Linear Differential Equation

 The first-order linear DE of the standard form y' + p(x)y = q(x), where p(x) and q(x) are continuous, is solved by the integrating-factor technique as follows (you don't have to remember things in brackets that start with “thus”):     Make sure that the DE is in the standard form y' + p(x) y = q(x), where the coefficient of y’ is 1, because the above technique assumes the standard form. If it's not in the standard form, convert it to the standard form.

##### Example 1.1

In Section 7.5 Part 3, we solved the DE dy/dt = ky, or y' + by = 0, where b = – k, and found the general solution to be y =
Cekt, where C = y(0). Use the integrating-factor technique described in this section to show that the general solution of the DE:

y' + by = 0,

where y = y(x) is a function of x and b is a constant, is y = y(0)ebx. Note that we've dealt with this DE under the heading “A
Particular Case” above.

## Note

Here the functions p(x) = b and q(x) = 0 are constant and thus continuous.

Solution
An antiderivative of
b is bx. Then:

ebxy’ + ebxby = 0     (multiply both sides of the given DE by ebx),
(
ebxy)’ = 0,

y(x) = y = Cebx,
y(0) = Ce0 = C(1) = C,
y = y(0)ebx.
EOS

##### Remark 1.1 - On Example 1.1

Let's find out what happens if we utilize the general antiderivative bx + c of b, where c is an arbitrary constant, instead of the
simple particular antiderivative
bx, where c = 0:

ebx+cy’ + ebx+cby = 0     (multiply both sides of the given DE by ebx+c),
(
ebx+cy)’ = 0,

y(x) = y = Ce-bx-c = Ce-ce-bx,
y(0) = Ce-ce0 = Ce-c(1) = Ce-c,
y = y(0)ebx.

It's the same general solution. Thus utilizing any particular antiderivative of p(x) is fine. We utilize the simplest one, where
the constant of integration is 0.

##### Example 1.2

Solve y' + y/x = 2, where x > 0.

Solution

EOS

##### Example 1.3

Solve the first-order linear DE dy/dx + xy = 2x3.

Solution

EOS

Example 1.4

Solve this initial-value problem:

Solution

EOS

 2. Applications

We now present examples of some of the applications of the first-order linear DEs.

##### Example 2.1

A tank contains 200 L of brine in which 2.5 kg of salt is dissolved. Then brine containing 0.1 kg of salt per liter enters the tank at
a rate of 15 L/min. While the brine is entering the tank, brine from the tank is also exiting the tank at a rate of 8 L/min. The
concentration of the brine in the tank is kept uniform throughout the tank by stirring. Find the amount of salt in the tank after 30
min.

Solution
Let
S be the amount of salt in the tank at time t, where S is in kg and t in min. The initial time t = 0 is when the brine starts to
enter and exit the tank. The rate of change of the amount of salt in the tank at time
t is:

C = (-17.5)(2008/7),

After 30 min, the amount of salt in the tank is approximately 33.3 kg.

EOS

##### Remark 2.1 - On Example 2.1

We're to determine a quantity. First we find the rate of change or derivative of that quantity. We obtain a DE whose unknown
function is the quantity. Then we solve the DE and get the quantity.

##### Example 2.2

A savings account is opened with a deposit of \$10,000. Money is being continuously transferred automatically from another
account and deposited into this account in such a way that the accumulated deposited amount increases at the rate of (1,000 +
200
t) dollars per year, where t is the number of years after the account is opened. Interest is being paid also continuously into
the account at the rate of 13% per year. Determine the balance of the account after 5 years.

## Notes

1. The quantity 1,000 + 200t is the rate of increase of the accumulated deposited amount (ADA), not of the deposit, at time t.
For example, for the simpler case of an ADA increasing at the constant rate of \$5,000 per year, the deposit is constant at
\$5,000 per year and thus increases at the rate of \$0 per year.

2. Since the interest is paid continuously, the accumulated interest (AI) at time t increases at (13%)B = \$0.13B per year, where

B is the balance at time t. In this problem the unit of time is the year. So the rate of increase of the AI (not the rate of
interest, not the rate of increase of the rate of interest, the rate of interest being fixed at 13% per year) is also \$0.13
B per
year.

3. Since the deposit is made continuously (the number of deposits on any time interval approaches infinity), we wonder if the
ADA is infinite and thus the balance is infinite at any time
t > 0 (the initial time t = 0 is when the account is opened). Let's
find out. As the rate of increase of the ADA is 1,000 + 200
t and the initial amount (at time t = 0) is \$10,000, the ADA at
time
t is a(t) = 1,000t + 200(t2/2) + 10,000, or a(t) = 100t2 + 1,000t + 10,000. So at any time t where 0 < t < infinity, the
a(t) is finite. Although the parabola a(t) rises or increases continuously for all increasing t > 0, its height at any t
where 0 <
t < infinity is finite. For any bounded interval [t1, t2] where 0 < t1 < t2, divide [t1, t2] into n equal sub-intervals,
and consider the parabola as increasing
n times on [t1, t2] from a(t1) to a(t2). For the continuous increase, let n approach
infinity. As the number of increases approaches infinity, each increase approaches 0 and fast, fast enough for their sum to be
finite. Thus the height of the parabola at any
t where 0 < t < infinity is finite. Although the deposits are made continuously,
each is small enough that the ADA is finite. As for the interest, we already know that although it's paid continuously, the
accumulated interest is finite. Thus at any time
t where 0 < t < infinity the balance is finite.

Solution
Let
B be the balance at time t. Then the rate of increase of the balance at time t is:

Let u = t and dv = e-0.13t dt, so that du = dt and v = -e-0.13t/0.13. Then:

The balance of the account after 5 years is \$29,340.53.
EOS

# Problems & Solutions

1. Solve the DE y' – 2y/x = x2.

##### Solution

An antiderivative of –2/x is –2 ln x. Then:

2. Solve y' + y = 3ex.

##### Solution

An antiderivative of 1 is x, and so an integrating factor is ex. We have:

Solution

y' + (cot x) y – 2x – 1 = 0,
y' + (cot x) y = 2x + 1.

4. Solve this initial-value problem:

Solution

(1 + sin x) y' + (cos x) y = sin2 x,

5. A tank contains 240 L of brine in which 6 kg of salt is dissolved. Then brine containing 0.06 kg of salt per litre from outside
the tank flows into it at a rate of 24 L/min and the brine in it flows out of it at a rate of 12 L/min. The concentration of brine
in the tank is kept uniform by stirring. Determine the amount of salt in the tank after 10 minutes.

Solution

Let S be the amount of salt in the tank at time t, where S is in kg and t in min. The initial time t = 0 is when the brine starts to
flow into and out of the tank. The rate of change of the amount of salt in the tank at time
t is:

After 10 min, the amount of salt in the tank is 16 kg.

6. The initial balance of a savings account was \$2,000. Interest is paid into the account continuously at the variable rate of (1 +
(
t/60))% per month, where t is the number of months after the account was opened. Determine the amount in the account
after 1 year.

Solution

Let a(t) be the amount in the account t months after the account was opened. The rate of increase of a(t) is:

2,000 = a(0) = Ce0 = C,

After 1 year or 12 months, the amount in the account is \$2,282.22

7. The equation y' + p(x)y = q(x)yn, where n is a constant different from 0 and 1, is known as Bernoulli equation. (For n = 0,
the equation becomes
y' + p(x)y = q(x), and for n = 1, it becomes y' + (p(x) – q(x))y = 0. In both cases, the equation is
of first-order and linear, and can be solved by the integrating-factor technique discussed in this section.)

a. Prove that the substitution v = y1–n reduces the Bernoulli equation to a DE for v that is of first-order and linear.
b. Use the result of part a to solve the DE
y' – 2xy = 5xy3.

##### Solution

a. Let v = y1–n. Then: