Differential Equations · Lesson 5 of 6

Linear Equations and Integrating Factors

“Use the product rule to derive integrating factors and solve linear equations in either variable.”

Learning Objectives

• Recognise a first-order linear equation and put it into standard form. • Identify the coefficient and forcing function using the correct independent variable. • Derive the integrating factor from the product rule. • Solve general and particular linear equations with clear integration steps. • Treat x as a function of y when that makes the equation linear. • Check intervals, initial conditions, and the meaning of a stated slope.

A different route to integration

Separation works when variable-dependent factors can be separated completely. A linear equation has a different useful structure: the dependent variable and its first derivative occur to the first power, with coefficients depending only on the independent variable. Multiplying by a carefully chosen function turns its left-hand side into one derivative.

Definition

First-order linear equation

An equation is linear in $y (x)$ if it can be written as $\frac{d y}{d x} + P (x) y = Q (x)$ , where $P$ and $Q$ are functions of $x$ only, including constants.

Linear standard formLaTeX

\frac{d y}{d x} + P (x) y = Q (x)

y

is the dependent variable.

P (x)

is its coefficient after the coefficient of

d y / d x

has been made one, and

Q (x)

is the remaining function of the independent variable.

For $y^{'} + y = sin x$ , we have $P = 1$ and $Q = sin x$ . For $y^{'} + y / x = e^{x}$ , $P = 1/ x$ and $Q = e^{x}$ . The equation $y^{'} + y^{2} = x$ is not linear in $y$ , and neither is $y^{'} + sin y = x$ : these involve nonlinear functions of the dependent variable.

You must first normalise the coefficient of the derivative. For $x y^{'} + 2 y = x^{2}$ , divide by $x$ on an interval where $x \neq = 0$ to get $y^{'} + (2/ x) y = x$ . Reading $P = 2$ from the unnormalised equation would give the wrong integrating factor.

Why the integrating factor works

The product rule tells us how to combine a derivative and a multiple of the original function. We want $g (x) y^{'} + P (x) g (x) y$ to equal the derivative of $g (x) y$ . Choosing the multiplier to make this happen gives the method its explanation.

The product rule gives $\frac{d}{d x} [g y] = g y^{'} + g^{'} y$ . Therefore we require $g^{'} = P g$ . A nonzero solution satisfies $g^{'} / g = P$ . Integrating gives $ln ∣ g ∣ = \int P d x$ plus a constant; one convenient positive choice is $g = e^{\int P d x}$ .

Integrating factorLaTeX

μ (x) = e^{\int P (x) d x}

The symbol

μ

denotes a multiplier depending only on

x

. The arbitrary multiplicative constant can be omitted because any nonzero constant multiple works equally well.

Now multiply the entire standard-form equation by $μ$ . The left side becomes $μ y^{'} + μ^{'} y$ , which is $d (μ y) / d x$ . Integrating both sides gives a direct expression for the product $μ y$ .

Integrated linear equationLaTeX

μ y = \int μ Q d x + C

After evaluating the integral, divide by the nonzero integrating factor if an explicit expression for

y

is wanted. Use an initial condition to determine

C

afterward.

For a logarithmic integral, use absolute values as needed: $\int d x / x = ln ∣ x ∣$ . An integrating factor may be simplified on a fixed interval. For instance, $e^{2 l n ∣ x ∣} = x^{2}$ , and a convenient multiplier for $P = - 1/ x$ is $1/ x$ on either interval $x > 0$ or $x < 0$ ; it differs from $e^{- l n ∣ x ∣}$ only by a constant sign there.

Example — A constant coefficient with a trigonometric term

Problem
Solve $y^{'} - y = cos x$ .

1.Here $P = - 1$ and $Q = cos x$ , so choose $μ = e^{- x}$ .
2.Multiplication gives $e^{- x} y^{'} - e^{- x} y = e^{- x} cos x$ , or $\frac{d}{d x} (e^{- x} y) = e^{- x} cos x$ .
3.Thus $e^{- x} y = I + C$ , where $I = \int e^{- x} cos x d x$ . Integrating by parts gives $I = - e^{- x} cos x - \int e^{- x} sin x d x$ .
4.Another integration by parts gives $\int e^{- x} sin x d x = - e^{- x} sin x + I$ . Hence $I = - e^{- x} cos x + e^{- x} sin x - I$ .
5.Therefore $I = \frac{1}{2} e^{- x} (sin x - cos x)$ . Divide by $e^{- x}$ to obtain $y = \frac{1}{2} (sin x - cos x) + C e^{x}$ .
6.The integrating factor simplified the differential equation; ordinary integration techniques are still needed for the right-hand side.

Example — Normalise before finding the multiplier

Problem
Solve $x y^{'} + 2 y = x^{2}$ , with $x \neq = 0$ .

1.Divide by $x$ : $y^{'} + (2/ x) y = x$ . Now $P = 2/ x$ , $Q = x$ , and $μ = e^{2 l n ∣ x ∣} = x^{2}$ .
2.Multiply through: $x^{2} y^{'} + 2 x y = x^{3}$ . The left side is $d (x^{2} y) / d x$ .
3.Integrate: $x^{2} y = x^{4} /4 + C$ , giving $y = x^{2} /4 + C / x^{2}$ .
4.The result applies separately on intervals that avoid zero. Substituting the derivative into $x y^{'} + 2 y$ cancels the constant terms and leaves $x^{2}$ .

Example — A logarithmic coefficient

Problem
Solve $x ln x y^{'} + y = \frac{2}{x} ln x$ on an interval where $x > 0$ and $x \neq = 1$ .

1.Divide by $x ln x$ : $y^{'} + \frac{1}{x l n x} y = 2/ x^{2}$ .
2.The integral of $P$ is $ln ∣ ln x ∣$ . A convenient integrating factor is $ln x$ , nonzero on the chosen interval.
3.Then $\frac{d}{d x} (y ln x) = 2 ln x / x^{2}$ . Integration by parts gives $\int 2 ln x / x^{2} d x = - 2 (ln x + 1) / x$ .
4.Thus $y ln x = - 2 (ln x + 1) / x + C$ . Division by $ln x$ gives an explicit solution if desired.

Linear in the other variable

Linearity depends on which variable is treated as dependent. A relation that is nonlinear as an equation for $y (x)$ may be linear as an equation for $x (y)$ . Solve for the derivative in both directions when the first form looks unhelpful.

Definition

Reverse linear form

An equation linear in $x (y)$ has the form $\frac{d x}{d y} + P_{1} (y) x = Q_{1} (y)$ . Its coefficient functions depend only on $y$ .

Reverse-variable integrating factorLaTeX

ν (y) = e^{\int P_{1} (y) d y}, ν x = \int ν Q_{1} d y + C

The integration variable is

y

throughout. The product to differentiate is

ν (y) x (y)

Example — A relation linear in x

Problem
Solve $y d x - (x + 2 y^{2}) d y = 0$ .

1.For $y \neq = 0$ , divide by $y d y$ to get $d x / d y - x / y = 2 y$ . It is linear in $x$ , with $P_{1} = - 1/ y$ and $Q_{1} = 2 y$ .
2.Choose $ν = 1/ y$ on a nonzero-y interval. Then $\frac{d}{d y} (x / y) = 2$ .
3.Integrate: $x / y = 2 y + C$ , giving $x = 2 y^{2} + C y$ .
4.Check directly: $d x / d y = 4 y + C$ , so $y d x / d y - (x + 2 y^{2}) = y (4 y + C) - (4 y^{2} + C y) = 0$ .
5.The derivation used $y \neq = 0$ . The curve $y = 0$ also satisfies the original differential relation when read as a constant function of $x$ ; it is not represented by using $y$ as an independent coordinate.

Example — An inverse-tangent integrating factor

Problem
Solve $(tan^{- 1} y - x) d y = (1 + y^{2}) d x$ .

1.Write $d x / d y + x / (1 + y^{2}) = tan^{- 1} y / (1 + y^{2})$ . This is linear in $x$ .
2.Here $P_{1} = 1/ (1 + y^{2})$ , so $ν = e^{t a n^{- 1} y}$ .
3.The solution is $x e^{t a n^{- 1} y} = \int \frac{t a n ^{- 1} y}{1 + y ^{2}} e^{t a n^{- 1} y} d y + C$ .
4.Set $t = tan^{- 1} y$ , so $d t = d y / (1 + y^{2})$ . The integral becomes $\int t e^{t} d t = e^{t} (t - 1)$ .
5.Divide by $e^{t a n^{- 1} y}$ to obtain $x = tan^{- 1} y - 1 + C e^{- t a n^{- 1} y}$ .

Use initial conditions after the integral

An initial condition determines the remaining constant. It also helps choose an interval on which the coefficients and integrating factor are defined. Keep the product form until the condition has been substituted if that reduces algebra.

Example — An integrating factor involving sine

Problem
Solve $y^{'} + y cot x = 2 x + x^{2} cot x$ with $y (π /2) = 0$ .

1.On an interval avoiding the zeros of sine, $P = cot x$ . Choose $μ = sin x$ .
2.Multiplication gives $\frac{d}{d x} (y sin x) = 2 x sin x + x^{2} cos x$ .
3.The right-hand side is the derivative of $x^{2} sin x$ by the product rule. Thus $y sin x = x^{2} sin x + C$ .
4.At $x = π /2$ , $y = 0$ , so $C = - π^{2} /4$ . Therefore $y = x^{2} - \frac{π ^{2}}{4 s i n x}$ .
5.The natural coefficient interval containing the given point is $0 < x < π$ . The original equation is undefined at the endpoints, so the solution is considered on that interval.

Example — Translate a tangent-slope description

Problem
A curve passes through $(0, 1)$ . At $(x, y)$ its tangent slope equals x plus the product xy. Find the curve.

1.The statement means $y^{'} = x + x y$ . Rearrange as $y^{'} - x y = x$ , so $P = - x$ and $Q = x$ .
2.The integrating factor is $μ = e^{- x^{2} /2}$ . Therefore $y e^{- x^{2} /2} = \int x e^{- x^{2} /2} d x + C$ .
3.Use $u = - x^{2} /2$ , so $d u = - x d x$ . The integral is $- e^{- x^{2} /2}$ .
4.Hence $y = - 1 + C e^{x^{2} /2}$ . The point gives $1 = - 1 + C$ , so $C = 2$ .
5.The curve is $y = - 1 + 2 e^{x^{2} /2}$ . It is also separable as $y^{'} = x (1 + y)$ , illustrating that a valid equation may fit more than one method.

Example — A first-degree appearance is not enough

Problem
Is $y^{'} + x y = y^{2}$ linear in y? Compare it with $y^{'} + x y = x^{2}$ .

1.In the first equation, moving terms to the left leaves a $y^{2}$ term. It cannot be written with $P$ and $Q$ depending only on $x$ , so it is not linear in $y$ .
2.The second equation already has standard form, with $P = x$ and $Q = x^{2}$ .
3.Both equations contain the derivative only to power one. That makes them first degree as differential equations, but only the second is linear in the dependent variable.

Multiply every term

An integrating factor multiplies the derivative term, the dependent-variable term, and the right-hand side. Multiplying only the derivative term does not produce the product rule identity. After normalising, check $μ^{'} = P μ$ if you are unsure about a sign.

A verbal condition can also be inconsistent. “The sum of the coordinates exceeds the magnitude of the tangent slope by five” means $x + y = ∣ y^{'} ∣ + 5$ . At $(0, 2)$ this would require $2 = ∣ y^{'} ∣ + 5$ , which is impossible. Dropping “magnitude” and using a signed derivative creates a different mathematical problem; it should not be done silently.

Check your understanding

Identify the dependent variable, normalise the derivative coefficient, and read the coefficient function before choosing an integrating factor. The questions also test the distinction between degree and linearity.

Quick check

For $x y^{'} + 2 y = x^{2}$ , what is P after normalisation?

Quick check

For $y^{'} - y = cos x$ , which is a valid integrating factor?

Quick check

Which equation is linear in y?

Quick check

Which identity explains the integrating-factor method?

Quick check

For $d x / d y - x / y = 2 y$ , which multiplier is convenient on a nonzero-y interval?

Quick check

If $P = cot x$ , which is a convenient integrating factor where sine is nonzero?

Quick check

Which equation can be solved by either separation or a linear integrating factor?

Quick check

After finding $μ$ , which integrated relation is correct?

Practice Problems

Solve $y^{'} + 2 y = sin x$ .
Solve $y^{'} + 3 y = e^{- 2 x}$ .
Solve $y^{'} + y / x = x^{2}$ for $x \neq = 0$ .
Solve $y^{'} + (sec x) y = tan x$ on $0 \leq x < π /2$ .
Solve $cos^{2} x y^{'} + y = tan x$ on $0 \leq x < π /2$ .
Solve $x y^{'} + 2 y = x^{2} ln x$ for $x > 0$ .
Solve $(1 + x^{2}) y^{'} + 2 x y = cot x$ where sine is nonzero.
Solve $(x + y) y^{'} = 1$ by treating x as a function of y.
Solve $y d x + (x - y^{2}) d y = 0$ for $y \neq = 0$ .
Solve $(x + 3 y^{2}) y^{'} = y$ for $y > 0$ .
Solve $y^{'} + 2 y tan x = sin x$ with $y (π /3) = 0$ .
Solve $(1 + x^{2}) y^{'} + 2 x y = 1/ (1 + x^{2})$ with $y (1) = 0$ .
Find a curve through the origin with tangent slope equal to the sum of its coordinates.
Explain why no curve through $(0, 2)$ can satisfy $x + y = ∣ y^{'} ∣ + 5$ .
Find an integrating factor for $(1 - y^{2}) d x / d y + y x = a y$ on $- 1 < y < 1$ .
Solve $x y^{'} + y - x + x y cot x = 0$ on an interval where $x sin x \neq = 0$ .
Solve $y^{'} - 3 y cot x = sin 2 x$ with $y (π /2) = 2$ .
Find an integrating factor and solve $x y^{'} - y = 2 x^{2}$ for $x \neq = 0$ .

$y = (2 sin x - cos x) /5 + C e^{- 2 x}$ . Use $μ = e^{2 x}$ and integration by parts.

Key Takeaways

• Write a linear equation with derivative coefficient one before identifying its coefficient function. • The integrating factor is chosen so that the left side becomes the derivative of a product. • For $y^{'} + P y = Q$ , use $μ = e^{\int P d x}$ and $μ y = \int μ Q d x + C$ . • If the equation is linear in $x (y)$ , integrate with respect to $y$ throughout. • Apply given conditions only on intervals where the original equation is defined, and check that verbal conditions are consistent.

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Homogeneous Equations and Ratio Substitutions

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Chapter Summary and Practice

A different route to integration

Why the integrating factor works

Linear in the other variable

Use initial conditions after the integral

Check your understanding

Quiz

Practice Problems

Key Takeaways