First-order Linear Equations

Earlier, we studied an application of a first-order differential equation that involved solving for the velocity of an object. In particular, if a ball is thrown upward with an initial velocity of

v_{0}

ft/s, then an initial-value problem that describes the velocity of the ball after

t

This model assumes that the only force acting on the ball is gravity. Now we add to the problem by allowing for the possibility of air resistance acting on the ball.

Air resistance always acts in the direction opposite to motion. Therefore if an object is rising, air resistance acts in a downward direction. If the object is falling, air resistance acts in an upward direction ([link]). There is no exact relationship between the velocity of an object and the air resistance acting on it. For very small objects, air resistance is proportional to velocity; that is, the force due to air resistance is numerically equal to some constant

k

For larger (e.g., baseball-sized) objects, depending on the shape, air resistance can be approximately proportional to the square of the velocity. In fact, air resistance may be proportional to

v^{1.5},

We will work with the linear approximation for air resistance. If we assume

k > 0,

Therefore the sum of the forces acting on the object is equal to the sum of the gravitational force and the force due to air resistance. This, in turn, is equal to the mass of the object multiplied by its acceleration at time

t

is the initial velocity measured in meters per second. This makes

g = 9.8 {m/s}^{2} .

The differential equation in this initial-value problem is an example of a first-order linear differential equation. (Recall that a differential equation is first-order if the highest-order derivative that appears in the equation is

1.)

In this section, we study first-order linear equations and examine a method for finding a general solution to these types of equations, as well as solving initial-value problems involving them.

is the dependent variable. Some examples of first-order linear differential equations are

etc. Due to these terms, it is impossible to put these equations into the same form as [link].

Standard Form

Our main goal in this section is to derive a solution method for equations of this form. It is useful to have the coefficient of

y^{'}

This is called the standard form of the differential equation. We will use it later when finding the solution to a general first-order linear differential equation. Returning to [link], we can divide both sides of the equation by

a (x) .

We can write any first-order linear differential equation in this form, and this is referred to as the standard form for a first-order linear differential equation.

Writing First-Order Linear Equations in Standard Form

Put each of the following first-order linear differential equations into standard form. Identify $p (x)$

and $q (x)$

for each equation.

$y' = 3 x - 4 y$
$\frac{3 x y'}{4 y - 3} = 2$
(here
$x > 0)$
$y = 3 y' - 4 x^{2} + 5$

Add $4 y$
to both sides:

$y' + 4 y = 3 x .$

In this equation,
$p (x) = 4$
and
$q (x) = 3 x .$
Multiply both sides by $4 y - 3,$
then subtract
$8 y$
from each side:

$\begin{matrix} \frac{3 x y'}{4 y - 3} & = & 2 \\ 3 x y' & = & 2 (4 y - 3) \\ 3 x y' & = & 8 y - 6 \\ 3 x y' - 8 y & = & −6 . \end{matrix}$

Finally, divide both sides by
$3 x$
to make the coefficient of
$y'$
equal to
$1 :$

$y' - \frac{8}{3 x} y = - \frac{2}{3 x} .$

This is allowable because in the original statement of this problem we assumed that
$x > 0 .$
(If
$x = 0$
then the original equation becomes
$0 = 2,$
which is clearly a false statement.)

In this equation,
$p (x) = - \frac{8}{3 x}$
and
$q (x) = - \frac{2}{3 x} .$
Subtract $y$
from each side and add
$4 x^{2} - 5 :$

$3 y' - y = 4 x^{2} - 5 .$

Next divide both sides by
$3 :$

$y' - \frac{1}{3} y = \frac{4}{3} x^{2} - \frac{5}{3} .$

In this equation,
$p (x) = - \frac{1}{3}$
and
$q (x) = \frac{4}{3} x^{2} - \frac{5}{3} .$

Integrating Factors

We now develop a solution technique for any first-order linear differential equation. We start with the standard form of a first-order linear differential equation:

The first term on the left-hand side of [link] is the derivative of the unknown function, and the second term is the product of a known function with the unknown function. This is somewhat reminiscent of the power rule from the Differentiation Rules section. If we multiply [link] by a yet-to-be-determined function

μ (x),

because it appears in the differential equation we are solving. Separating variables and integrating yields

can be an arbitrary (positive or negative) constant. This leads to a general method for solving a first-order linear differential equation. We first multiply both sides of [link] by the integrating factor

μ (x) .

was previously calculated, we are now finished. An important note about the integrating constant

C :

It may seem that we are inconsistent in the usage of the integrating constant. However, the integral involving

p (x)

is necessary in order to find an integrating factor for [link]. Only one integrating factor is needed in order to solve the equation; therefore, it is safe to assign a value for

C

When calculating the integral inside the brackets in [link], it is necessary to keep our options open for the value of the integrating constant, because our goal is to find a general family of solutions to [link]. This integrating factor guarantees just that.

Solving a First-order Linear Equation

Find a general solution for the differential equation $x y' + 3 y = 4 x^{2} - 3 x .$

Assume $x > 0 .$

To put this differential equation into standard form, divide both sides by $x :$

$y' + \frac{3}{x} y = 4 x - 3 .$

Therefore
$p (x) = \frac{3}{x}$
and
$q (x) = 4 x - 3 .$
The integrating factor is $μ (x) = e^{\int (3 / x) d x} = e^{3 ln x} = x^{3} .$
Multiplying both sides of the differential equation by $μ (x)$
gives us

$\begin{matrix} x^{3} y^{'} + x^{3} (\frac{3}{x}) y & = & x^{3} (4 x - 3) \\ x^{3} y^{'} + 3 x^{2} y & = & 4 x^{4} - 3 x^{3} \\ \frac{d}{d x} (x^{3} y) & = & 4 x^{4} - 3 x^{3} . \end{matrix}$
Integrate both sides of the equation.

$\begin{matrix} \int \frac{d}{d x} (x^{3} y) d x & = & \int 4 x^{4} - 3 x^{3} d x \\ x^{3} y & = & \frac{4 x^{5}}{5} - \frac{3 x^{4}}{4} + C \\ y & = & \frac{4 x^{2}}{5} - \frac{3 x}{4} + C x^{−3} . \end{matrix}$
There is no initial value, so the problem is complete.

Analysis

You may have noticed the condition that was imposed on the differential equation; namely, $x > 0 .$

For any nonzero value of $C,$

the general solution is not defined at $x = 0 .$

Furthermore, when $x < 0,$

the integrating factor changes. The integrating factor is given by [link] as $f (x) = e^{\int p (x) d x} .$

For this $p (x)$

we get

e^{\int p (x) d x =} e^{\int (3 / x) d x} = e^{3 ln \| x \|} = {\| x \|}^{3},

since $x < 0 .$

The behavior of the general solution changes at $x = 0$

largely due to the fact that $p (x)$

is not defined there.

A First-order Linear Initial-Value Problem

Solve the initial-value problem

y^{'} + 3 y = 2 x - 1, y (0) = 3 .

This differential equation is already in standard form with $p (x) = 3$
and
$q (x) = 2 x - 1 .$
The integrating factor is $μ (x) = e^{\int 3 d x} = e^{3 x} .$
Multiplying both sides of the differential equation by $μ (x)$
gives

$\begin{matrix} e^{3 x} y^{'} + 3 e^{3 x} y & = & (2 x - 1) e^{3 x} \\ \frac{d}{d x} [y e^{3 x}] & = & (2 x - 1) e^{3 x} . \end{matrix}$

Integrate both sides of the equation:

$\begin{matrix} \int \frac{d}{d x} [y e^{3 x}] d x & = & \int (2 x - 1) e^{3 x} d x \\ y e^{3 x} & = & \frac{e^{3 x}}{3} (2 x - 1) - \int \frac{2}{3} e^{3 x} d x \\ y e^{3 x} & = & \frac{e^{3 x} (2 x - 1)}{3} - \frac{2 e^{3 x}}{9} + C \\ y & = & \frac{2 x - 1}{3} - \frac{2}{9} + C e^{−3 x} \\ y & = & \frac{2 x}{3} - \frac{5}{9} + C e^{−3 x} . \end{matrix}$
Now substitute $x = 0$
and
$y = 3$
into the general solution and solve for
$C :$

$\begin{matrix} y & = & \frac{2}{3} x - \frac{5}{9} + C e^{−3 x} \\ 3 & = & \frac{2}{3} (0) - \frac{5}{9} + C e^{−3 (0)} \\ 3 & = & - \frac{5}{9} + C \\ C & = & \frac{32}{9} . \end{matrix}$

Therefore the solution to the initial-value problem is

$y = \frac{2}{3} x - \frac{5}{9} + \frac{32}{9} e^{−3 x} .$

Applications of First-order Linear Differential Equations

We look at two different applications of first-order linear differential equations. The first involves air resistance as it relates to objects that are rising or falling; the second involves an electrical circuit. Other applications are numerous, but most are solved in a similar fashion.

Free fall with air resistance

We discussed air resistance at the beginning of this section. The next example shows how to apply this concept for a ball in vertical motion. Other factors can affect the force of air resistance, such as the size and shape of the object, but we ignore them here.

A Ball with Air Resistance

A racquetball is hit straight upward with an initial velocity of $2$

m/s. The mass of a racquetball is approximately $0.0427$

kg. Air resistance acts on the ball with a force numerically equal to $0.5 v,$

where $v$

represents the velocity of the ball at time $t .$

Find the velocity of the ball as a function of time.
How long does it take for the ball to reach its maximum height?
If the ball is hit from an initial height of $1$
meter, how high will it reach?

The mass $m = 0.0427 kg, k = 0.5,$
and
$g = 9.8 {m/s}^{2} .$
The initial velocity is
$v_{0} = 2$
m/s. Therefore the initial-value problem is

$0.0427 \frac{d v}{d t} = −0.5 v - 0.0427 (9.8), v_{0} = 2 .$

Dividing the differential equation by
$0.0427$
gives

$\frac{d v}{d t} = −11.7096 v - 9.8, v_{0} = 2 .$

The differential equation is linear. Using the problem-solving strategy for linear differential equations:

Step 1. Rewrite the differential equation as
$\frac{d v}{d t} + 11.7096 v = −9.8 .$
This gives
$p (t) = 11.7096$
and
$q (t) = −9.8$

Step 2. The integrating factor is
$μ (t) = e^{\int 11.7096 d t} = e^{11.7096 t} .$

Step 3. Multiply the differential equation by
$μ (t) :$

$\begin{matrix} e^{11.7096 t} \frac{d v}{d t} + 11.7096 v e^{11.7096 t} & = & −9.8 e^{11.7096 t} \\ \frac{d}{d t} [v e^{11.7096 t}] & = & −9.8 e^{11.7096 t} . \end{matrix}$

Step 4. Integrate both sides:

$\begin{matrix} \int \frac{d}{d t} [v e^{11.7096 t}] d t & = & \int −9.8 e^{11.7096 t} d t \\ v e^{11.7096 t} & = & \frac{−9.8}{11.7096} e^{11.7096 t} + C \\ v (t) & = & −0.8369 + C e^{−11.7096 t} . \end{matrix}$

Step 5. Solve for
$C$
using the initial condition
$v_{0} = v (0) = 2 :$

$\begin{matrix} v (t) & = & −0.8369 + C e^{−11.7096 t} \\ v (0) & = & −0.8369 + C e^{−11.7096 (0)} \\ 2 & = & −0.8369 + C \\ C & = & 2.8369. \end{matrix}$

Therefore the solution to the initial-value problem is
$v (t) = 2.8369 e^{−11.7096 t} - 0.8369 .$
The ball reaches its maximum height when the velocity is equal to zero. The reason is that when the velocity is positive, it is rising, and when it is negative, it is falling. Therefore when it is zero, it is neither rising nor falling, and is at its maximum height:

$\begin{matrix} 2.8369 e^{−11.7096 t} - 0.8369 & = & 0 \\ 2.8369 e^{−11.7096 t} & = & 0.8369 \\ e^{−11.7096 t} & = & \frac{0.8369}{2.8369} \approx 0.295 \\ ln e^{−11.7096 t} & = & ln 0.295 \approx - 1.221 \\ −11.7096 t & = & −1.221 \\ t & \approx & 0.104. \end{matrix}$

Therefore it takes approximately
$0.104$
second to reach maximum height.
To find the height of the ball as a function of time, use the fact that the derivative of position is velocity, i.e., if $h (t)$
represents the height at time
$t,$
then
$h^{'} (t) = v (t) .$
Because we know
$v (t)$
and the initial height, we can form an initial-value problem:

$h^{'} (t) = 2.8369 e^{−11.7096 t} - 0.8369, h (0) = 1 .$

Integrating both sides of the differential equation with respect to
$t$
gives

$\begin{matrix} \int h^{'} (t) d t & = & \int 2.8369 e^{−11.7096 t} - 0.8369 d t \\ h (t) & = & - \frac{2.8369}{11.7096} e^{−11.7096 t} - 0.8369 t + C \\ h (t) & = & −0.2423 e^{−11.7096 t} - 0.8369 t + C . \end{matrix}$

Solve for
$C$
by using the initial condition:

$\begin{matrix} h (t) & = & −0.2423 e^{−11.7096 t} - 0.8369 t + C \\ h (0) & = & −0.2423 e^{−11.7096 (0)} - 0.8369 (0) + C \\ 1 & = & −0.2423 + C \\ C & = & 1.2423. \end{matrix}$

Therefore

$h (t) = −0.2423 e^{−11.7096 t} - 0.8369 t + 1.2423 .$

After
$0.104$
second, the height is given by

$h (0.2) = −0.2423 e^{−11.7096 t} - 0.8369 t + 1.2423 \approx 1.0836$
meter.

Electrical Circuits

A source of electromotive force (e.g., a battery or generator) produces a flow of current in a closed circuit, and this current produces a voltage drop across each resistor, inductor, and capacitor in the circuit. Kirchhoff’s Loop Rule states that the sum of the voltage drops across resistors, inductors, and capacitors is equal to the total electromotive force in a closed circuit. We have the following three results:

Next, suppose there is no inductor in the circuit, but there is a capacitor and a resistor, so

L = 0, R \neq 0,

which is a first-order linear differential equation. This is referred to as an RC circuit. In either case, we can set up and solve an initial-value problem.

Finding Current in an *RL* Electric Circuit

A circuit has in series an electromotive force given by $E = 50 sin 20 t V,$

a resistor of $5 Ω,$

and an inductor of $0.4 H .$

If the initial current is $0,$

find the current at time $t > 0 .$

We have a resistor and an inductor in the circuit, so we use [link]. The voltage drop across the resistor is given by $E_{R} = R i = 5 i .$

The voltage drop across the inductor is given by $E_{L} = L i^{'} = 0.4 i^{'} .$

The electromotive force becomes the right-hand side of [link]. Therefore [link] becomes

0.4 i^{'} + 5 i = 50 sin 20 t .

Dividing both sides by $0.4$

gives the equation

i^{'} + 12.5 i = 125 sin 20 t .

Since the initial current is 0, this result gives an initial condition of $i (0) = 0 .$

We can solve this initial-value problem using the five-step strategy for solving first-order differential equations.

Step 1. Rewrite the differential equation as $i^{'} + 12.5 i = 125 sin 20 t .$

This gives $p (t) = 12.5$

and $q (t) = 125 sin 20 t .$

Step 2. The integrating factor is $μ (t) = e^{\int 12.5 d t} = e^{12.5 t} .$

Step 3. Multiply the differential equation by $μ (t) :$

\begin{matrix} e^{12.5 t} i^{'} + 12.5 e^{12.5 t} i & = & 125 e^{12.5 t} sin 20 t \\ \frac{d}{d t} [i e^{12.5 t}] & = & 125 e^{12.5 t} sin 20 t . \end{matrix}

Step 4. Integrate both sides:

\begin{matrix} \int \frac{d}{d t} [i e^{12.5 t}] d t & = & \int 125 e^{12.5 t} sin 20 t d t \\ i e^{12.5 t} & = & (\frac{250 sin 20 t - 400 cos 20 t}{89}) e^{12.5 t} + C \\ i (t) & = & \frac{250 sin 20 t - 400 cos 20 t}{89} + C e^{−12.5 t} . \end{matrix}

Step 5. Solve for $C$

using the initial condition $v (0) = 2 :$

\begin{matrix} i (t) & = & \frac{250 sin 20 t - 400 cos 20 t}{89} + C e^{−12.5 t} \\ i (0) & = & \frac{250 sin 20 (0) - 400 cos 20 (0)}{89} + C e^{−12.5 (0)} \\ 0 & = & - \frac{400}{89} + C \\ C & = & \frac{400}{89} . \end{matrix}

Therefore the solution to the initial-value problem is $i (t) = \frac{250 sin 20 t - 400 cos 20 t + 400 e^{−12.5 t}}{89} = \frac{250 sin 20 t - 400 cos 20 t}{89} + \frac{400 e^{−12.5 t}}{89} .$

The first term can be rewritten as a single cosine function. First, multiply and divide by $\sqrt{250^{2} + 400^{2}} = 50 \sqrt{89} :$

\begin{matrix} \frac{250 sin 20 t - 400 cos 20 t}{89} & = \frac{50 \sqrt{89}}{89} (\frac{250 sin 20 t - 400 cos 20 t}{50 \sqrt{89}}) \\ = - \frac{50 \sqrt{89}}{89} (\frac{8 cos 20 t}{\sqrt{89}} - \frac{5 sin 20 t}{\sqrt{89}}) . \end{matrix}

Next, define $φ$

to be an acute angle such that $cos φ = \frac{8}{\sqrt{89}} .$

Then $sin φ = \frac{5}{\sqrt{89}}$

and

\begin{matrix} - \frac{50 \sqrt{89}}{89} (\frac{8 cos 20 t}{\sqrt{89}} - \frac{5 sin 20 t}{\sqrt{89}}) & = - \frac{50 \sqrt{89}}{89} (cos φ cos 20 t - sin φ sin 20 t) \\ = - \frac{50 \sqrt{89}}{89} cos (20 t + φ) . \end{matrix}

Therefore the solution can be written as

i (t) = - \frac{50 \sqrt{89}}{89} cos (20 t + φ) + \frac{400 e^{−12.5 t}}{89} .

The second term is called the attenuation term, because it disappears rapidly as t grows larger. The phase shift is given by $φ,$

and the amplitude of the steady-state current is given by $\frac{50 \sqrt{89}}{89} .$

The graph of this solution appears in [link]:

A graph of the given solution over [0, 6] on the x axis. It is an oscillating function, rapidly going from just below -5 to just above 5.

Key Concepts

Key Equations

Are the following differential equations linear? Explain your reasoning.

\frac{d y}{d x} = x^{2} y + sin x

\frac{d y}{d t} = t y

Yes

\frac{d y}{d t} + y^{2} = x

y' = x^{3} + e^{x}

Yes

y' = y + e^{y}

Write the following first-order differential equations in standard form.

y' = x^{3} y + sin x

y' - x^{3} y = sin x

y' + 3 y - ln x = 0

− x y' = (3 x + 2) y + x e^{x}

y' + \frac{(3 x + 2)}{x} y = − e^{x}

\frac{d y}{d t} = 4 y + t y + tan t

\frac{d y}{d t} = y x (x + 1)

\frac{d y}{d t} - y x (x + 1) = 0

What are the integrating factors for the following differential equations?

y' = x y + 3

y' + e^{x} y = sin x

e^{x}

y' = x ln (x) y + 3 x

\frac{d y}{d x} = tanh (x) y + 1

− ln (cosh x)

\frac{d y}{d t} + 3 t y = e^{t} y

Solve the following differential equations by using integrating factors.

y' = 3 y + 2

y = C e^{3 x} - \frac{2}{3}

y' = 2 y - x^{2}

x y' = 3 y - 6 x^{2}

y = C x^{3} + 6 x^{2}

(x + 2) y' = 3 x + y

y' = 3 x + x y

y = C e^{x^{2} / 2} - 3

x y' = x + y

sin (x) y' = y + 2 x

y = C tan (\frac{x}{2}) - 2 x + 4 tan (\frac{x}{2}) ln (sin (\frac{x}{2}))

y' = y + e^{x}

x y' = 3 y + x^{2}

y = C x^{3} - x^{2}

y' + ln x = \frac{y}{x}

Solve the following differential equations. Use your calculator to draw a family of solutions. Are there certain initial conditions that change the behavior of the solution?

[T] $(x + 2) y' = 2 y - 1$

y = C {(x + 2)}^{2} + \frac{1}{2}

[T] $y' = 3 e^{t / 3} - 2 y$

[T] $x y' + \frac{y}{2} = sin (3 t)$

y = \frac{C}{\sqrt{x}} + 2 sin (3 t)

[T] $x y' = 2 \frac{cos x}{x} - 3 y$

[T] $(x + 1) y' = 3 y + x^{2} + 2 x + 1$

y = C {(x + 1)}^{3} - x^{2} - 2 x - 1

[T] $sin (x) y' + cos (x) y = 2 x$

[T] $\sqrt{x^{2} + 1} y' = y + 2$

y = C e^{{sinh}^{−1} x} - 2

[T] $x^{3} y' + 2 x^{2} y = x + 1$

Solve the following initial-value problems by using integrating factors.

y' + y = x, y (0) = 3

y = x + 4 e^{x} - 1

y' = y + 2 x^{2}, y (0) = 0

x y' = y - 3 x^{3}, y (1) = 0

y = - \frac{3 x}{2} (x^{2} - 1)

x^{2} y' = x y - ln x, y (1) = 1

(1 + x^{2}) y' = y - 1, y (0) = 0

y = 1 - e^{{tan}^{−1} x}

x y' = y + 2 x ln x, y (1) = 5

(2 + x) y' = y + 2 + x, y (0) = 0

y = (x + 2) ln (\frac{x + 2}{2})

y' = x y + 2 x e^{x}, y (0) = 2

\sqrt{x} y' = y + 2 x, y (0) = 1

y = 2 e^{2 \sqrt{x}} - 2 x - 2 \sqrt{x} - 1

y' = 2 y + x e^{x}, y (0) = −1

A falling object of mass $m$

can reach terminal velocity when the drag force is proportional to its velocity, with proportionality constant $k .$

Set up the differential equation and solve for the velocity given an initial velocity of $0 .$

v (t) = \frac{g m}{k} (1 - e^{− k t / m})

Using your expression from the preceding problem, what is the terminal velocity? (Hint: Examine the limiting behavior; does the velocity approach a value?)

[T] Using your equation for terminal velocity, solve for the distance fallen. How long does it take to fall $5000$

meters if the mass is $100$

kilograms, the acceleration due to gravity is $9.8$

m/s² and the proportionality constant is $4 ?$

40.451

seconds

A more accurate way to describe terminal velocity is that the drag force is proportional to the square of velocity, with a proportionality constant $k .$

Set up the differential equation and solve for the velocity.

Using your expression from the preceding problem, what is the terminal velocity? (Hint: Examine the limiting behavior: Does the velocity approach a value?)

\sqrt{\frac{g m}{k}}

[T] Using your equation for terminal velocity, solve for the distance fallen. How long does it take to fall $5000$

meters if the mass is $100$

kilograms, the acceleration due to gravity is $9.8 {m/s}^{2}$

and the proportionality constant is $4 ?$

Does it take more or less time than your initial estimate?

For the following problems, determine how parameter $a$

affects the solution.

Solve the generic equation $y' = a x + y .$

How does varying $a$

change the behavior?

y = C e^{x} - a (x + 1)

Solve the generic equation $y' = a x + y .$

How does varying $a$

change the behavior?

Solve the generic equation $y' = a x + x y .$

How does varying $a$

change the behavior?

y = C e^{x^{2} / 2} - a

Solve the generic equation $y' = x + a x y .$

How does varying $a$

change the behavior?

Solve $y' - y = e^{k t}$

with the initial condition $y (0) = 0 .$

As $k$

approaches $1,$

what happens to your formula?

y = \frac{e^{k t} - e^{t}}{k - 1}

Chapter Review Exercises

For the following problems, find the general solution to the differential equations.

For the following problems, draw the directional field associated with the differential equation, then solve the differential equation. Draw a sample solution on the directional field.

Then solve the initial-value problem exactly. How close is your Euler’s Method estimate?

First-order Linear Equations

Standard Form

Integrating Factors

Applications of First-order Linear Differential Equations

Free fall with air resistance

Electrical Circuits

Key Concepts

Key Equations

Chapter Review Exercises

Glossary