Chapter 2 · Section 2.3

2.3 Linear Equations and Bernoulli Equations

Linear first-order equations are everywhere in engineering. The current in an RL circuit, the charge on a capacitor, the temperature of a cooling part, the concentration in a stirred tank: each obeys an equation of the form "rate of change plus a multiple of the quantity equals a source". The good news is that every such equation can be solved by one method. In this section we derive that method carefully, and then use it to solve a nonlinear family, the Bernoulli equations.

Linear first-order equations

Definition 2.3.1 (Linear first-order equation)

A first-order equation is linear in $y$ if it can be written as

a_{1} (x) y^{'} + a_{0} (x) y = b (x) .

On an interval where $a_{1} (x) \neq = 0$ , dividing by $a_{1}$ gives the standard form

y^{'} + P (x) y = Q (x) .

"Linear" means that $y$ and $y^{'}$ appear only to the first power, are not multiplied together, and are not inside other functions. The coefficients may depend on $x$ in any way. For example,

$x y^{'} + 2 y = 5 x^{3}$ is linear; its standard form is $y^{'} + \frac{2}{x} y = 5 x^{2}$ (for $x \neq = 0$ );
$y^{'} + y^{2} = x$ is not linear (because of $y^{2}$ );
$y^{'} + sin y = 0$ is not linear (because of $sin y$ ).

Common mistake

Always divide by the coefficient of $y^{'}$ before reading off $P$ and $Q$ . For $x y^{'} + 2 y = 5 x^{3}$ , the coefficient $P$ is $2/ x$ , not $2$ .

Deriving the integrating factor

The left side $y^{'} + P y$ looks almost like the result of the product rule. The idea is to multiply the equation by a function $μ (x)$ that turns it into an exact derivative. Let us find such a $μ$ in small steps.

Multiply the standard form by an unknown positive function $μ (x)$ : $μ y^{'} + μ P y = μ Q .$
Recall the product rule. $(μ y)^{'} = μ y^{'} + μ^{'} y .$
Compare. The left side in step 1 equals $(μ y)^{'}$ exactly when the coefficients of $y$ agree: $μ^{'} = μ P .$
Solve for $μ$ . This is a separable equation for $μ$ . Divide by $μ > 0$ : $\frac{μ ^{'}}{μ} = P (x) ⟹ ln μ = \int P (x) d x .$
Exponentiate. $μ (x) = e^{\int P (x) d x} .$ Any antiderivative will do, so we take the constant of integration equal to zero.

With this choice the equation becomes

(μ (x) y)^{'} = μ (x) Q (x) .

Now integrate both sides and divide by $μ$ (which is never zero):

μ (x) y = \int μ (x) Q (x) d x + c ⟹ y = \frac{1}{μ ( x )} [\int μ (x) Q (x) d x + c] .

The same factor from exactness

Write the equation in differential form: $[P (x) y - Q (x)] d x + d y = 0$ . Here $M_{y} = P$ and $N_{x} = 0$ , so it is not exact (unless $P = 0$ ). Multiply by $μ (x)$ . The test of Section 2.1 now asks $\partial_{y} (μ (P y - Q)) = \partial_{x} μ$ , i.e. $μ P = μ^{'}$ . That is the same equation as in step 3. So $e^{\int P d x}$ is precisely an integrating factor in the sense of Section 2.1.

Two details that cause trouble

The constant inside $\int P d x$ . Why may we drop it? If we kept it, we would get $μ = e^{\int P d x + k} = e^{k} e^{\int P d x}$ . That just multiplies the whole equation by the nonzero number $e^{k}$ , which cancels when we divide by $μ$ at the end. So any antiderivative of $P$ gives the same final answer. Pick the simplest one.

The constant after integrating $μ Q$ . Here the constant matters a great deal. It must be added before dividing by $μ$ .

Common mistake

Writing $y = \frac{1}{μ} \int μ Q d x + c$ is wrong. The correct form is $y = \frac{1}{μ} [\int μ Q d x + c]$ , so the constant is also divided by $μ$ . In Example 2.3.1 below this is the difference between the correct $c e^{- 4 x}$ and an incorrect $+ c$ , which does not even satisfy the equation.

The structure of the answer

Split the general solution into two pieces:

y = y_{h} c e^{- \int P d x} + y_{p} e^{- \int P d x} \int e^{\int P d x} Q d x .

The first piece $y_{h}$ solves the equation with $Q = 0$ (the "unforced" equation). The second piece $y_{p}$ is one particular solution of the full equation. "General solution = unforced solution + one particular solution" is a pattern you will see again for all linear equations in Chapter 4.

Existence and uniqueness

For linear equations, the theory is especially clean. Because every step above can be reversed, the formula gives all solutions, not just some.

Theorem 2.3.1 (Linear initial-value problems)

Let $P (x)$ and $Q (x)$ be continuous on an open interval $I$ , let $x_{0} \in I$ , and let $y_{0}$ be any real number. Then the initial-value problem

y^{'} + P (x) y = Q (x), y (x_{0}) = y_{0}

has exactly one solution, and this solution exists on the whole interval $I$ . With $μ (x) = exp (\int_{x_{0}}^{x} P (s) d s)$ it is

y (x) = \frac{1}{μ ( x )} [y_{0} + \int_{x_{0}}^{x} μ (s) Q (s) d s] .

Compare this with nonlinear equations, whose solutions can blow up at points you cannot see in the equation (Example 2.2.4 was one). A linear equation can only misbehave where $P$ or $Q$ does.

Method in a nutshell

Put the equation in standard form $y^{'} + P (x) y = Q (x)$ . Note the intervals where $P$ and $Q$ are continuous.
Compute $μ = e^{\int P d x}$ (no constant needed). Simplify, e.g. $e^{2 l n x} = x^{2}$ .
Multiply through by $μ$ and write the left side as $(μ y)^{'}$ .
Integrate: $μ y = \int μ Q d x + c$ . Do not forget $c$ .
Divide by $μ$ . Use the initial condition, if any, to find $c$ .

Practise with the solver below. In the "const" mode $P$ is a constant; in the "power" mode $P (x) = k / x$ , as in the Cauchy–Euler-like equation $x y^{'} + k y = x Q (x)$ . It shows $μ$ , the integral $\int μ Q d x$ , and the final answer.

Worked examples: linear equations

Example 2.3.1

Solve $y^{'} + 4 y = 10 e^{x}$ .

Solution

Standard form. Already: $P = 4$ , $Q = 10 e^{x}$ , both continuous on $R$ .
Integrating factor. $μ = e^{\int 4 d x} = e^{4 x}$ .
Multiply. $e^{4 x} y^{'} + 4 e^{4 x} y = 10 e^{5 x}$ , i.e. $(e^{4 x} y)^{'} = 10 e^{5 x}$ .
Integrate. $e^{4 x} y = 2 e^{5 x} + c$ .
Divide. $y = 2 e^{x} + c e^{- 4 x}$ .

As $x \to \infty$ , the term $c e^{- 4 x}$ dies out and every solution approaches $2 e^{x}$ , whatever the starting value.

Figure 2.3.1. Solutions

y = 2 e^{x} + c e^{- 4 x}

of Example 2.3.1 for six values of

c

. The term

c e^{- 4 x}

dies out quickly, and every solution merges with the particular solution

2 e^{x}

Example 2.3.2

Solve $x y^{'} + 2 y = 5 x^{3}$ .

Solution

Standard form. Divide by $x$ (so $x \neq = 0$ ): $y^{'} + \frac{2}{x} y = 5 x^{2}$ . We work on $x > 0$ or on $x < 0$ .
Integrating factor. $\int \frac{2}{x} d x = 2 ln ∣ x ∣$ , so $μ = e^{2 l n ∣ x ∣} = x^{2}$ .
Multiply. $x^{2} y^{'} + 2 x y = 5 x^{4}$ , i.e. $(x^{2} y)^{'} = 5 x^{4}$ .
Integrate. $x^{2} y = x^{5} + c$ .
Divide. $y = x^{3} + \frac{c}{x ^{2}}$ .

To try this kind of equation in the solver, choose the "power" mode with $k = 2$ and $Q = 5 x^{2}$ .

Example 2.3.3

Solve $y^{'} + (tan x) y = sin 2 x$ , $y (0) = 3$ .

Solution

Interval. $tan x$ is continuous on $- \frac{π}{2} < x < \frac{π}{2}$ , which contains $x_{0} = 0$ . By Theorem 2.3.1 the solution exists on this whole interval.
Integrating factor. $\int tan x d x = - ln (cos x)$ there (since $cos x > 0$ ), so $μ = e^{- l n c o s x} = \frac{1}{cos x} = sec x .$
Multiply. Using $sin 2 x = 2 sin x cos x$ : $(sec x \cdot y)^{'} = \frac{2 sin x cos x}{cos x} = 2 sin x$ .
Integrate. $sec x \cdot y = - 2 cos x + c$ .
Divide (multiply by $cos x$ ): $y = - 2 cos^{2} x + c cos x$ .
Initial condition. $y (0) = - 2 + c = 3$ , so $c = 5$ and $y = 5 cos x - 2 cos^{2} x, - \frac{π}{2} < x < \frac{π}{2} .$

Example 2.3.4 (an RL circuit)

A circuit has an inductance $L = 1$ H, a resistance $R = 2 Ω$ and a voltage source $E (t) = 10 sin t$ V. By Kirchhoff's voltage law the current $i (t)$ satisfies $L i^{'} + R i = E (t)$ . The switch closes at $t = 0$ , so $i (0) = 0$ . Find $i (t)$ .

Solution

Equation. $i^{'} + 2 i = 10 sin t$ , with $P = 2$ .
Integrating factor. $μ = e^{2 t}$ , giving $(e^{2 t} i)^{'} = 10 e^{2 t} sin t$ .
The integral. Integrating by parts twice (or using the standard formula $\int e^{a t} sin b t d t = \frac{e ^{a t} ( a sin b t - b cos b t )}{a ^{2} + b ^{2}}$ with $a = 2$ , $b = 1$ ): $\int 10 e^{2 t} sin t d t = 2 e^{2 t} (2 sin t - cos t) + c .$
Divide by $e^{2 t}$ . $i = 4 sin t - 2 cos t + c e^{- 2 t}$ .
Initial condition. $i (0) = - 2 + c = 0$ , so $c = 2$ : $i (t) = steady state 4 sin t - 2 cos t + transient 2 e^{- 2 t} .$
Interpretation. The transient decays with time constant $L / R = 0.5$ s. After a few seconds the current is a pure sinusoid with amplitude $4^{2} + 2^{2} = 25 \approx 4.47$ A, at the same frequency as the source.

Figure 2.3.2. Current in the RL circuit of Example 2.3.4. The transient

2 e^{- 2 t}

is practically gone after about

2.5

s (five time constants

L / R

). From then on the current coincides with the steady-state sinusoid of amplitude

25 \approx 4.47

Example 2.3.5 (linear in

x

instead of

y

)

Solve $y^{'} = \frac{y}{2 x + y ^{3}}$ .

Solution

Not linear in $y$ . Because of $y^{3}$ in the denominator, no rearrangement makes this linear in $y$ .
Swap roles. Regard $x$ as a function of $y$ . Since $\frac{d x}{d y} = 1 / \frac{d y}{d x}$ (where $y^{'} \neq = 0$ ), $\frac{d x}{d y} = \frac{2 x + y ^{3}}{y} = \frac{2}{y} x + y^{2} .$ This is linear in $x$ : $\frac{d x}{d y} - \frac{2}{y} x = y^{2}$ .
Integrating factor. $μ = e^{\int - 2/ y d y} = e^{- 2 l n ∣ y ∣} = y^{- 2}$ .
Multiply and integrate. $\frac{d}{d y} (\frac{x}{y ^{2}}) = 1$ , so $\frac{x}{y ^{2}} = y + c$ .
Answer. $x = y^{3} + c y^{2}$ .
Lost solution. We needed $y^{'} \neq = 0$ to flip the derivative. The constant $y \equiv 0$ gives $y^{'} = 0$ and $\frac{0}{2 x} = 0$ (for $x \neq = 0$ ), so $y \equiv 0$ is also a solution.

Long-term behaviour: transient and steady state

Examples 2.3.1 and 2.3.4 show a pattern that engineers rely on. When $P$ is a positive constant, the unforced part of the solution is $c e^{- P x}$ . It decays to zero, whatever the value of $c$ . So, after a while, every solution looks like the particular solution $y_{p}$ , and the initial condition is "forgotten".

The decaying piece is called the transient, and what remains is the steady state (or forced response). The number $1/ P$ is the time constant: after one time constant the transient has shrunk by the factor $e^{- 1} \approx 0.37$ , and after five time constants by $e^{- 5} \approx 0.007$ . In practice, an RL circuit is said to reach steady state after about five time constants $L / R$ , and an RC circuit after about five times $RC$ .

If $P$ is negative, the opposite happens: $c e^{- P x}$ grows, and small changes in the initial value are amplified. Exercise 2 is an example. Recognising the sign of $P$ tells you, before any integration, whether the system is stable.

Bernoulli equations

Some nonlinear equations are linear equations in disguise. The most important family is named after Jacob Bernoulli.

Definition 2.3.2 (Bernoulli equation)

An equation of the form

y^{'} + P (x) y = Q (x) y^{n}

where $n$ is a real number, is called a Bernoulli equation.

If $n = 0$ or $n = 1$ , the equation is already linear. For any other $n$ , the term $y^{n}$ makes it nonlinear. The logistic population model $y^{'} = ry - s y^{2}$ is a Bernoulli equation with $n = 2$ .

The substitution $v = y^{1 - n}$

Divide by $y^{n}$ (assume $y \neq = 0$ ): $y^{- n} y^{'} + P (x) y^{1 - n} = Q (x) .$
Name the new unknown. Let $v = y^{1 - n}$ . By the chain rule, $v^{'} = (1 - n) y^{- n} y^{'} ⟹ y^{- n} y^{'} = \frac{v ^{'}}{1 - n} .$
Substitute. $\frac{1}{1 - n} v^{'} + P (x) v = Q (x) .$
Multiply by $(1 - n)$ . $v^{'} + (1 - n) P (x) v = (1 - n) Q (x) .$ This is linear in $v$ . Solve it with the integrating factor, then return to $y = v^{1/ (1 - n)}$ .

Do not forget

y \equiv 0

When $n > 0$ , the function $y \equiv 0$ solves the Bernoulli equation ( $0 + 0 = 0$ ). Step 1 divided by $y^{n}$ , so this solution is usually lost. Always add it to your answer.

Two more remarks on the exponent.

If $n < 0$ , then $y \equiv 0$ is not a solution, because $y^{n}$ is undefined at $y = 0$ . For example, $y^{'} + y = 3 y^{- 2}$ only makes sense where $y \neq = 0$ .
If $n$ is not an integer, $y^{n}$ is usually defined only for $y > 0$ (for $n = \frac{1}{2}$ , say). Then we look for positive solutions, and the substitution $v = y^{1 - n}$ is safe.

Why does the substitution work at all? The term that made the equation nonlinear, $Q y^{n}$ , became the "source" $Q$ after dividing by $y^{n}$ . At the same time, the left side $y^{- n} y^{'} + P y^{1 - n}$ turned out to be a linear expression in the single quantity $v = y^{1 - n}$ and its derivative. The power $1 - n$ is chosen precisely so that its derivative produces the factor $y^{- n} y^{'}$ .

Method in a nutshell

Write the equation as $y^{'} + P y = Q y^{n}$ and read off $n$ .
Set $v = y^{1 - n}$ . The new equation is $v^{'} + (1 - n) P v = (1 - n) Q$ .
Solve this linear equation for $v$ .
Return to $y$ using $y^{1 - n} = v$ .
Add $y \equiv 0$ if $n > 0$ .

Use the solver to explore Bernoulli equations with constant $P$ and $Q$ . Try different exponents $n$ , including negative and fractional ones.

Worked examples: Bernoulli equations

Example 2.3.6

Solve $y^{'} + 2 y = 4 y^{3}$ .

Solution

Identify. Bernoulli with $P = 2$ , $Q = 4$ , $n = 3$ .
Substitute. $v = y^{1 - 3} = y^{- 2}$ . Then $1 - n = - 2$ , so $v^{'} + (- 2) (2) v = (- 2) (4) ⟹ v^{'} - 4 v = - 8.$
Solve for $v$ . $μ = e^{- 4 x}$ : $(e^{- 4 x} v)^{'} = - 8 e^{- 4 x} ⟹ e^{- 4 x} v = 2 e^{- 4 x} + c ⟹ v = 2 + c e^{4 x} .$
Back to $y$ . $y^{- 2} = 2 + c e^{4 x}$ , so $y^{2} = \frac{1}{2 + c e ^{4 x}} or y = \pm \frac{1}{2 + c e ^{4 x}} .$
Zero solution. $n = 3 > 0$ , so $y \equiv 0$ is also a solution.

Note that the constant solutions $y = \pm 1/ 2$ correspond to $c = 0$ . They are the equilibria where $2 y = 4 y^{3}$ .

Figure 2.3.3. Solutions

y = \pm 1/ 2 + c e^{4 x}

of Example 2.3.6. Far to the left every curve is close to one of the equilibria

\pm 1/ 2

. Curves with

c > 0

then settle on the zero solution; curves with

c < 0

blow up where

2 + c e^{4 x} = 0

Example 2.3.7

Solve $y^{'} + \frac{2}{x} y = - x y^{2}$ , $y (1) = \frac{1}{2}$ .

Solution

Identify. Bernoulli with $P = 2/ x$ , $Q = - x$ , $n = 2$ . Work on $x > 0$ .
Substitute. $v = y^{- 1}$ and $1 - n = - 1$ : $v^{'} - \frac{2}{x} v = x .$
Integrating factor. $μ = e^{- 2 l n x} = x^{- 2}$ , so $(\frac{v}{x ^{2}})^{'} = x \cdot x^{- 2} = \frac{1}{x} .$
Integrate. $\frac{v}{x ^{2}} = ln x + c$ , so $v = x^{2} (ln x + c)$ .
Back to $y$ . $y = \frac{1}{x ^{2} ( ln x + c )}$ .
Initial condition. $y (1) = \frac{1}{c} = \frac{1}{2}$ gives $c = 2$ : $y = \frac{1}{x ^{2} ( 2 + ln x )} .$
Interval. The denominator vanishes at $ln x = - 2$ , i.e. $x = e^{- 2}$ . The solution exists on $e^{- 2} < x < \infty$ . Unlike linear equations, Bernoulli equations can blow up at points that the coefficients do not reveal.

Exercises

In Exercises 1–6, find the general solution.

$y^{'} + 5 y = 15$
Answer
$y = 3 + c e^{- 5 x}$ .
$y^{'} - 2 y = 4 x$
Answer
$y = - 2 x - 1 + c e^{2 x}$ .
$y^{'} - 2 x y = 4 x$
Answer
$y = - 2 + c e^{x^{2}}$ .
$x y^{'} - y = x^{3}$ , $x > 0$
Answer
$y = \frac{1}{2} x^{3} + c x$ .
$y^{'} + (cot x) y = 2 cos x$ , $0 < x < π$
Answer
$y = sin x + c csc x$ .
$(x^{2} + 4) y^{'} + 4 x y = 6 x$
Answer
$y = \frac{3}{2} + \frac{c}{( x ^{2} + 4 ) ^{2}}$ .
An RC circuit with $R = 10 Ω$ , $C = 0.01$ F and a battery of $12$ V obeys $R q^{'} + q / C = 12$ , where $q$ is the charge in coulombs. If $q (0) = 0$ , find $q (t)$ and the final charge.
Answer
$10 q^{'} + 100 q = 12$ ; $q = 0.12 (1 - e^{- 10 t})$ ; final charge $0.12$ C.
Solve $x y^{'} + 3 y = 2 x^{2}$ , $y (1) = 1$ .
Answer
$y = \frac{2 x ^{2}}{5} + \frac{3}{5 x ^{3}}$ , for $x > 0$ .
Solve $y^{'} = \frac{1}{x + y ^{2}}$ by treating $x$ as a function of $y$ .
Answer
$x = c e^{y} - y^{2} - 2 y - 2$ .
Solve the Bernoulli equation $y^{'} + y = e^{x} y^{2}$ .
Answer
$y = \frac{e ^{- x}}{c - x}$ , and $y \equiv 0$ .
Solve $y^{'} + \frac{1}{x} y = 3 x^{2} y^{3}$ , $x > 0$ .
Answer
$y^{2} = \frac{1}{c x ^{2} - 6 x ^{3}}$ , and $y \equiv 0$ .
A population model reads $y^{'} = y - 3 y^{2}$ with $y (0) = \frac{1}{4}$ . Solve it and find $lim_{x \to \infty} y (x)$ .
Answer
$y = \frac{1}{3 + e ^{- x}}$ ; the limit is $\frac{1}{3}$ .

Linear first-order equations

Deriving the integrating factor

Two details that cause trouble

The structure of the answer

Existence and uniqueness

Worked examples: linear equations

Long-term behaviour: transient and steady state

Bernoulli equations

The substitution v=y1−n

Worked examples: Bernoulli equations

Exercises

The substitution $v = y^{1 - n}$