A Bernoulli differential equation has the form y’ + p(x)y = q(x)y^n, where n is any real number other than 0 or 1. The key to solving it is a substitution that transforms it into a standard linear equation you already know how to handle. Once you see the pattern, the process is straightforward and repeatable.
Recognizing the Bernoulli Form
The standard Bernoulli equation looks like this:
y’ + p(x)y = q(x)y^n
Here, p(x) and q(x) are continuous functions of x, and n is a real number. The y^n term on the right side is what makes this equation nonlinear and distinguishes it from an ordinary first-order linear ODE. If n = 0, the y^n term becomes 1 and the equation is already linear. If n = 1, the y^n term is just y, and you can move it to the left side to get a linear equation. So the Bernoulli method only applies when n is something other than 0 or 1, like n = 2, n = 3, n = -1, or even a fraction.
Before doing anything else, make sure your equation is in this exact form. You may need to divide through by a coefficient on y’ or rearrange terms so the y’ stands alone on one side with the y^n term isolated on the other.
The Substitution That Makes It Work
The entire strategy rests on one substitution:
v = y^(1-n)
This choice is not arbitrary. It’s specifically designed to cancel the nonlinear y^n term during the transformation. When you differentiate both sides with respect to x, you get:
dv/dx = (1 – n) y^(-n) dy/dx
This relationship lets you replace dy/dx in the original equation with an expression involving dv/dx. The y terms that remain will simplify cleanly because of how the exponents interact with v = y^(1-n).
Step-by-Step Solution Process
Step 1: Divide by y^n
Start with y’ + p(x)y = q(x)y^n and divide every term by y^n:
y^(-n) y’ + p(x) y^(1-n) = q(x)
This is the critical move. The left side now contains y^(-n) y’ and y^(1-n), both of which relate directly to your substitution variable v.
Step 2: Substitute v = y^(1-n)
Since v = y^(1-n), you know that y^(1-n) in the equation is simply v. And since dv/dx = (1 – n) y^(-n) y’, you can write y^(-n) y’ = (1/(1-n)) dv/dx. Plugging both into the equation gives:
(1/(1-n)) dv/dx + p(x) v = q(x)
Multiply through by (1 – n) to clean it up:
dv/dx + (1 – n) p(x) v = (1 – n) q(x)
Step 3: Solve the Linear Equation
You now have a first-order linear ODE in v, which you can solve using an integrating factor. For a linear equation of the form v’ + P(x)v = Q(x), the integrating factor is:
I(x) = e^(∫P(x) dx)
In this case, P(x) = (1 – n) p(x). Multiply both sides of the equation by I(x), recognize the left side as the derivative of I(x)·v, then integrate both sides with respect to x. The general solution for v is:
v(x) = e^(-∫P(x) dx) [ ∫ e^(∫P(x) dx) · (1 – n) q(x) dx + C ]
where C is the constant of integration.
Step 4: Back-Substitute for y
Once you have v(x), replace v with y^(1-n) and solve for y. Since v = y^(1-n), you get:
y = v^(1/(1-n))
This gives your final solution in terms of the original variable y. If you have an initial condition, plug it in now to find the constant C.
Worked Example: n = 2
Consider the equation y’ + y = y^2. Here p(x) = 1, q(x) = 1, and n = 2.
The substitution is v = y^(1-2) = y^(-1), so v = 1/y. The derivative gives dv/dx = -y^(-2) y’. Dividing the original equation by y^2 yields y^(-2) y’ + y^(-1) = 1, which becomes -dv/dx + v = 1, or equivalently dv/dx – v = -1.
This is a standard linear ODE. The integrating factor is e^(∫-1 dx) = e^(-x). Multiplying through:
d/dx [e^(-x) v] = -e^(-x)
Integrating both sides: e^(-x) v = e^(-x) + C, so v = 1 + Ce^(x).
Back-substituting v = 1/y gives y = 1/(1 + Ce^(x)). You can verify this works by differentiating and plugging back into the original equation.
Common Mistakes to Avoid
The most frequent error is forgetting to divide by y^n before substituting. If you try to substitute v = y^(1-n) directly into the original equation without this step, the exponents won’t simplify and you’ll end up with a mess. The division is what sets up the clean replacement.
Another common pitfall is losing track of the (1 – n) factor. When you replace y^(-n) y’ with (1/(1-n)) dv/dx, you need to multiply through by (1 – n) afterward. Forgetting this leaves an incorrect coefficient on dv/dx, which throws off the integrating factor and the entire solution.
Watch for sign errors when n is greater than 1. For example, when n = 2, the factor (1 – n) is -1, which flips signs throughout the transformed equation. When n = 3, (1 – n) = -2. These negative values frequently cause algebraic mistakes if you’re not careful.
Finally, don’t forget that y = 0 is often a singular solution that gets lost during the process. Dividing by y^n at the start implicitly assumes y ≠ 0, so check whether y = 0 satisfies the original equation separately.
Why This Method Works
The substitution v = y^(1-n) is powerful because it’s tailored to the exponent that causes the nonlinearity. In the original equation, the y^n term prevents you from using standard linear techniques. The substitution absorbs that nonlinearity into the change of variables itself. After dividing by y^n, every remaining y term has the exponent (1 – n), which matches v exactly. The equation collapses into a linear form where the integrating factor method applies directly.
This is a general technique in differential equations: when a nonlinear equation has a specific structural pattern, a well-chosen substitution can linearize it. The Bernoulli equation is one of the cleanest examples of this principle, which is why it appears in virtually every introductory ODE course. The same strategic thinking applies to other nonlinear forms like Riccati equations and certain separable equations, where the right change of variable turns something intractable into something routine.