Bernoulli Equations and Substitution Methods

Core Idea

A Bernoulli equation has the form y' + P(x)y = Q(x)yⁿ, where n is a real number other than 0 or 1. It is nonlinear, but the substitution v = y^1-n turns it into the linear equation v' + (1 - n)P(x)v = (1 - n)Q(x), which an integrating factor solves. The broader lesson is that a well-chosen change of variable can turn an equation you cannot solve into one you can. The logistic equation is a familiar Bernoulli equation with n = 2.

How It's Best Learned

Derive the substitution once in general so students see why the exponent 1 - n is the right choice, then practice the steps on equations with n = 2, n = 3 and n = 1/2. Always check the final answer by substituting it back into the original equation. Point out the solution y = 0, which is easy to lose when dividing by yⁿ.

Common Misconceptions

• Bernoulli equations must be solved by separation of variables. (Most are not separable; the substitution v = y^1-n makes them linear instead.)
• The substitution is v = yⁿ. (The correct substitution is v = y^1-n, which makes the y^1-n term and the derivative term match.)
• Dividing by yⁿ loses nothing. (For n > 0, the constant solution y = 0 can be lost and must be checked separately.)
• The method works for n = 0 or n = 1. (Those cases are already linear and need no substitution; the formula v = y^1-n is useless when n = 1.)

Explainer

You can solve first-order linear equations y' + P(x)y = Q(x) with an integrating factor. Many models are slightly nonlinear, though, and one family is close enough to linear that a substitution rescues it. A Bernoulli equation, named for Jacob Bernoulli, has the form

y' + P(x)y = Q(x)yⁿ, with n ≠ 0, 1.

When n = 0 or n = 1 the equation is linear. Otherwise the yⁿ term makes it nonlinear.

The idea is to divide by yⁿ and look for a new variable. Dividing gives

y⁻ⁿy' + P(x)y^1-n = Q(x).

Now set v = y^1-n. By the chain rule, v' = (1 - n)y⁻ⁿy', so y⁻ⁿy' = v'/(1 - n). Substituting,

v'/(1 - n) + P(x)v = Q(x), or v' + (1 - n)P(x)v = (1 - n)Q(x).

This equation is linear in v. Solve it with an integrating factor, then recover y from y = v^(1/(1-n)). The exponent 1 - n is chosen exactly so that the derivative term and the P(x) term involve the same new variable.

Example: y' + y = y². Here n = 2, so v = y⁻¹. Dividing by y² gives y⁻²y' + y⁻¹ = 1. Since v' = -y⁻²y', this becomes -v' + v = 1, or v' - v = -1. The integrating factor is e⁻ˣ, and the solution is v = 1 + Ceˣ. Therefore y = 1/(1 + Ceˣ). Check by substitution: the derivative is y' = -Ceˣ/(1 + Ceˣ)², and y' + y simplifies to 1/(1 + Ceˣ)², which is y².

When we divided by y², we assumed y ≠ 0. The constant function y = 0 also solves y' + y = y², and it does not come from the formula for any value of C. For n > 0, always check y = 0 separately.

The logistic population model P' = rP(1 - P/K) is a Bernoulli equation with n = 2. The substitution v = 1/P turns it into the linear equation v' + rv = r/K, which gives a second route to the logistic solution besides separation of variables.

Bernoulli equations are one example of a general strategy. Homogeneous equations y' = F(y/x) yield to the substitution v = y/x, and equations y' = G(ax + by) yield to v = ax + by. In each case, the right substitution turns an unfamiliar equation into a known type.