Bernoulli's Equation

Bernoulli's equation looks like a linear equation spoiled by a power of $y$ on the right-hand side. The substitution $z = y^{1 - α}$ removes that power exactly, converting the equation into a linear one that the previous section already solves. This is the first of several places in the book where a well-chosen change of variable turns a hard problem into a solved one.

Quick Reference

Item	Statement
Bernoulli's equation	$\frac{d y}{d x} + P (x) y = Q (x) y^{α}$ , with $α$ real
Already linear when	$α = 0$ or $α = 1$
Preparatory step	Divide by $y^{α}$ : $y^{- α} \frac{d y}{d x} + P (x) y^{1 - α} = Q (x)$
Substitution	$z = y^{1 - α}$ , so $\frac{d z}{d x} = (1 - α) y^{- α} \frac{d y}{d x}$
Resulting linear equation	z' + (1-\alpha)P(x)z = (1-\alpha)Q(x)
Integrating factor	$I (x) = e^{(1 - α) \int P (x) d x}$
Back-substitute	$y = z^{1 / (1 - α)}$
Lost solution	$y = 0$ is a solution whenever $α > 0$ , and dividing by $y^{α}$ discards it

The Equation and Its Reduction

A Bernoulli equation is an equation of the form

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

where $α$ is a real number.

When $α = 0$ the right-hand side is just $Q (x)$ and the equation is already linear. When $α = 1$ the equation becomes y' + \left(P - Q\right)y = 0, again linear and in fact separable. So assume from here on that $α$ is neither $0$ nor $1$ .

Dividing both sides by $y^{α}$ , we have

y^{- α} \frac{d y}{d x} + P (x) y^{1 - α} = Q (x)

This is the crucial rearrangement. Notice that only two combinations of $y$ now appear: $y^{1 - α}$ in the middle term, and $y^{- α} \frac{d y}{d x}$ in the first term. The second is, up to a constant, the derivative of the first. That observation dictates the substitution.

The equation can be reduced to a linear equation by the change of variable

z = y^{1 - α} ⟹ \frac{d z}{d x} = (1 - α) y^{- α} \frac{d y}{d x}

so that $y^{- α} \frac{d y}{d x} = \frac{1}{1 - α} \frac{d z}{d x}$ . Substituting,

\frac{1}{1 - α} \frac{d z}{d x} + P (x) z = Q (x)

and multiplying through by $(1 - α)$ ,

z' + (1-\alpha)P(x)\,z = (1-\alpha)Q(x) .

This is a linear first-order equation in $z$ , solvable by the integrating factor $I (x) = e^{(1 - α) \int P d x}$ . Once $z$ is found, recover $y$ from $y = z^{1 / (1 - α)}$ .

Watch the sign and the factor. Both $P$ and $Q$ get multiplied by $(1 - α)$ , and when $α > 1$ that factor is negative, flipping the signs of both. For $α = 2$ , the most common case, $1 - α = - 1$ and the linear equation is z' - P(x)z = -Q(x).

Lost solution. Dividing by $y^{α}$ at the start is invalid where $y = 0$ . For $α > 0$ , the constant function $y = 0$ satisfies the original equation, and it is never recovered from the formula for $z$ , since $z = y^{1 - α}$ is undefined there. State it separately.

Worked Examples

Solve the initial value problem

y' + \frac{4}{x}y = x^{3}y^{2}, \qquad x > 0, \qquad y(1) = 1

and find the interval of validity for the solution.

Solution

This is a Bernoulli equation with

P (x) = \frac{4}{x}, Q (x) = x^{3}, α = 2 .

Since $1 - α = - 1$ , the substitution is $z = y^{- 1}$ , and the linear equation is

z' + (-1)\frac{4}{x}z = (-1)x^{3} \qquad \Longrightarrow \qquad z' - \frac{4}{x}z = -x^{3} .

The integrating factor is

I (x) = e^{\int (- 4 / x) d x} = e^{- 4 \ln x} = x^{- 4} (x > 0) .

Multiplying through,

\frac{d}{d x} (x^{- 4} z) = - x^{3} \cdot x^{- 4} = - \frac{1}{x} .

Integrating,

x^{- 4} z = - \ln x + c ⟹ z = x^{4} (c - \ln x) .

Back-substituting $z = 1 / y$ ,

y = \frac{1}{x^{4} (c - \ln x)} .

Apply the initial condition $y (1) = 1$ . At $x = 1$ we have $\ln 1 = 0$ , so

1 = \frac{1}{1 \cdot (c - 0)} = \frac{1}{c} ⟹ c = 1 .

The solution of the initial value problem is

y = \frac{1}{x^{4} (1 - \ln x)} .

Interval of validity. The solution is undefined where the denominator vanishes, that is, where $x = 0$ or $1 - \ln x = 0$ . The second gives $\ln x = 1$ , so $x = e$ . The initial point $x = 1$ lies between $0$ and $e$ , so the interval of validity is

0 < x < e .

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

Solution

Here $P = \frac{1}{x}$ , $Q = x$ , and $α = 2$ , so $z = y^{- 1}$ and

z' - \frac{1}{x}z = -x .

The integrating factor is $I = e^{- \int d x / x} = \frac{1}{x}$ , giving

\frac{d}{d x} (\frac{z}{x}) = - \frac{x}{x} = - 1 ⟹ \frac{z}{x} = - x + c ⟹ z = x (c - x) .

Therefore

y = \frac{1}{x (c - x)},

together with the separate solution $y = 0$ , which the division by $y^{2}$ discarded.

Solve the logistic equation $\frac{d P}{d t} = r P (1 - \frac{P}{K})$ as a Bernoulli equation, where $r$ and $K$ are positive constants.

Solution

Expand the right-hand side and move the linear term left:

\frac{d P}{d t} = r P - \frac{r}{K} P^{2} ⟹ \frac{d P}{d t} - r P = - \frac{r}{K} P^{2} .

This is Bernoulli with $α = 2$ , coefficient $- r$ in place of $P (x)$ , and forcing $- \frac{r}{K}$ . Substituting $z = P^{- 1}$ and multiplying by $1 - α = - 1$ ,

z' + rz = \frac{r}{K} .

The integrating factor is $e^{r t}$ , so

\frac{d}{d t} (z e^{r t}) = \frac{r}{K} e^{r t} ⟹ z e^{r t} = \frac{1}{K} e^{r t} + c ⟹ z = \frac{1}{K} + c e^{- r t} .

Therefore

P (t) = \frac{1}{\frac{1}{K} + c e^{- r t}} = \frac{K}{1 + c K e^{- r t}} .

As $t \to \infty$ the exponential dies and $P \to K$ , confirming the qualitative analysis of the direction field in Section 2.2: every positive solution approaches the carrying capacity.

Exercises

Solve y' + y = xy^{3}.

Solution

Here $α = 3$ , so $1 - α = - 2$ and $z = y^{- 2}$ . The linear equation is

z' - 2z = -2x .

With $I = e^{- 2 x}$ ,

\frac{d}{d x} (z e^{- 2 x}) = - 2 x e^{- 2 x} .

Integrating by parts, $\int - 2 x e^{- 2 x} d x = x e^{- 2 x} + \frac{1}{2} e^{- 2 x}$ , so

z e^{- 2 x} = x e^{- 2 x} + \frac{1}{2} e^{- 2 x} + c ⟹ z = x + \frac{1}{2} + c e^{2 x} .

Since $z = y^{- 2}$ ,

y^{2} = \frac{1}{x + \frac{1}{2} + c e^{2 x}},

plus the separate solution $y = 0$ .

Solve $x \frac{d y}{d x} + y = y^{2} \ln x$ for $x > 0$ .

Solution

Divide by $x$ to reach standard form:

\frac{d y}{d x} + \frac{1}{x} y = \frac{\ln x}{x} y^{2} .

With $α = 2$ , take $z = y^{- 1}$ and multiply by $- 1$ :

z' - \frac{1}{x}z = -\frac{\ln x}{x} .

The integrating factor is $1 / x$ , so

\frac{d}{d x} (\frac{z}{x}) = - \frac{\ln x}{x^{2}} .

Integrating by parts with $u = \ln x$ and $d v = - x^{- 2} d x$ , so $v = x^{- 1}$ :

\int - \frac{\ln x}{x^{2}} d x = \frac{\ln x}{x} - \int \frac{1}{x^{2}} d x = \frac{\ln x}{x} + \frac{1}{x} .

Hence $\frac{z}{x} = \frac{\ln x + 1}{x} + c$ , so $z = \ln x + 1 + c x$ and

y = \frac{1}{\ln x + 1 + c x} .

Explain why the substitution $z = y^{1 - α}$ fails when $α = 1$ , and solve y' + P(x)y = Q(x)y directly.

Solution

When $α = 1$ the substitution gives $z = y^{0} = 1$ , a constant, which carries no information. The derivative relation z' = (1-\alpha)y^{-\alpha}y' also collapses to $0 = 0$ .

But no substitution is needed. Collecting the $y$ terms,

y' + \left[P(x) - Q(x)\right]y = 0 ,

a homogeneous linear equation, and also separable. Its solution is

y = c \exp (- \int [P (x) - Q (x)] d x) .

Frequently Asked Questions

How do I recognize a Bernoulli equation?

It looks exactly like a linear equation except that the right-hand side carries a power of $y$ : y' + P(x)y = Q(x)y^{\alpha}. Everything else must be linear, so the coefficient of $y$ on the left may depend on $x$ alone, and there must be no other nonlinear appearance of $y$ .

Why does

z = y^{1 - α}

work?

After dividing by $y^{α}$ the equation contains only $y^{1 - α}$ and y^{-\alpha}y'. Differentiating $z = y^{1 - α}$ gives z' = (1-\alpha)y^{-\alpha}y', so the second expression is a constant multiple of z' and the first is $z$ itself. The substitution was reverse-engineered to make exactly this happen.

Must

α

be a whole number?

No. The derivation uses only the exponent rules, so $α$ may be any real number: $α = \frac{1}{2}$ , $α = - 3$ , and $α = π$ all work. Non-integer values do restrict the domain, since $y^{α}$ may require $y > 0$ .

What happened to the solution

y = 0

Dividing by $y^{α}$ threw it away. For $α > 0$ it is a genuine solution of the original equation, since both sides vanish. The transformed variable $z = y^{1 - α}$ cannot represent it, so add it back by hand when reporting the answer.

Is the logistic equation really a Bernoulli equation?

Yes, with $α = 2$ . It is also separable, and solving it by partial fractions is a common alternative. The Bernoulli route is shorter here, and it makes clear why the answer has the form of a constant over $1$ plus a decaying exponential.

Can I skip the substitution and just guess?

Not reliably. The substitution is mechanical and takes one line, whereas guessing the form of the answer for a general $P$ and $Q$ is hopeless. Do note that once you have practiced a few, you can go straight from the original equation to the linear equation z' + (1-\alpha)Pz = (1-\alpha)Q without writing the intermediate division step.

Quick Reference

The Equation and Its Reduction

Worked Examples

Exercises

Frequently Asked Questions

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