In short

Some nonlinear first-order ODEs become linear after a clever substitution. The most important example is Bernoulli's equation \dfrac{dy}{dx} + P(x)y = Q(x)y^n, which is linear only when n = 0 or n = 1. The substitution v = y^{1-n} turns it into a linear equation in v. After solving that linear equation, substitute back to get y. Several other nonlinear forms also yield to substitutions that turn them into linear or separable equations — the trick is always spot the hidden linear structure.

The logistic equation models population growth with a carrying capacity:

\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right)

Expand the right side: \frac{dP}{dt} = rP - \frac{r}{K}P^2. Bring everything with P to the left:

\frac{dP}{dt} - rP = -\frac{r}{K}P^2

This is not linear — there is a P^2 on the right. You cannot apply the integrating factor method. The variable-separable trick does work for this specific equation, but only because it is autonomous (no explicit t dependence). In general, an equation with a y^n on the right is not something you have a tool for yet.

Here is the observation that changes everything. If you substitute v = 1/P (that is, v = P^{-1}), then \frac{dv}{dt} = -P^{-2}\frac{dP}{dt}. Multiply the original equation by -P^{-2}:

-P^{-2}\frac{dP}{dt} + rP^{-1} = \frac{r}{K}

The left side is now \frac{dv}{dt} + rv. The equation has become

\frac{dv}{dt} + rv = \frac{r}{K}

This is a linear equation in v. You already know how to solve it — integrating factor e^{rt}, and so on. Once you have v(t), set P = 1/v and you are done.

What happened? A single substitution turned a nonlinear equation into a linear one. The trick is not restricted to the logistic equation — it applies to an entire family of equations called Bernoulli equations, and there are several other families with similar tricks. This article lays them out.

Bernoulli's equation

Bernoulli's Equation

A differential equation of the form

\frac{dy}{dx} + P(x)\,y = Q(x)\,y^n

where n is any real number (except 0 or 1, which give trivially linear equations), is called Bernoulli's equation.

Substitution: v = y^{1-n}. This turns the equation into a linear equation in v:

\frac{dv}{dx} + (1-n)\,P(x)\,v = (1-n)\,Q(x)

Solve for v(x) using the integrating factor method, then substitute back v = y^{1-n} to recover y.

Why the substitution works. Start with the Bernoulli equation and divide both sides by y^n:

y^{-n}\frac{dy}{dx} + P(x)\,y^{1-n} = Q(x)

Now set v = y^{1-n}. Then

\frac{dv}{dx} = (1 - n)\,y^{-n}\frac{dy}{dx}

So y^{-n}\frac{dy}{dx} = \frac{1}{1-n}\frac{dv}{dx}. Substitute into the equation:

\frac{1}{1-n}\frac{dv}{dx} + P(x)\,v = Q(x)

Multiply by (1-n):

\frac{dv}{dx} + (1-n)P(x)\,v = (1-n)Q(x)

This is a linear equation in v with coefficients (1-n)P and (1-n)Q. You solve it with the integrating factor e^{\int (1-n)P\,dx}, and then recover y from y = v^{1/(1-n)}.

Critical special values. If n = 0, the equation is already linear. If n = 1, the equation becomes \frac{dy}{dx} + (P - Q)y = 0, which is also linear (and separable). Only for n \neq 0, 1 does the Bernoulli substitution do anything useful.

The method at work

Solve \dfrac{dy}{dx} + \dfrac{y}{x} = x y^2.

Step 1. Recognise the form. This is y' + P y = Q y^n with P(x) = 1/x, Q(x) = x, and n = 2.

Step 2. Substitute v = y^{1-n} = y^{-1}. Then \frac{dv}{dx} = -y^{-2}\frac{dy}{dx}.

Step 3. Divide the original equation by y^2:

y^{-2}\frac{dy}{dx} + \frac{1}{x}\,y^{-1} = x

Substitute: the first term is -\frac{dv}{dx} and the second is \frac{v}{x}:

-\frac{dv}{dx} + \frac{v}{x} = x

Multiply by -1 to put it in standard form:

\frac{dv}{dx} - \frac{v}{x} = -x

Step 4. Solve the linear equation for v. Here P = -1/x, so

\mu(x) = e^{\int -1/x\,dx} = e^{-\ln|x|} = \frac{1}{x}

Multiply through:

\frac{1}{x}\frac{dv}{dx} - \frac{v}{x^2} = -1

The left side is \frac{d}{dx}(v/x), so

\frac{d}{dx}\left(\frac{v}{x}\right) = -1
\frac{v}{x} = -x + C
v = -x^2 + Cx

Step 5. Back-substitute. Since v = 1/y,

\frac{1}{y} = -x^2 + Cx
y = \frac{1}{Cx - x^2}

That is the general solution. At any particular initial condition, C is determined.

Three members of the solution family $y = \frac{1}{Cx - x^2}$ for the Bernoulli equation $y' + y/x = xy^2$. Each curve has two vertical asymptotes — the points where $Cx - x^2 = 0$, namely $x = 0$ and $x = C$. The curves live in the strip between these asymptotes; the solution cannot be extended past $x = C$.

Other nonlinear equations reducible to linear

Bernoulli is the most famous case, but it is not the only one. A few other nonlinear forms respond to specific substitutions.

Equations with y^2 on the right

The logistic equation \frac{dy}{dx} = ay - by^2 is a Bernoulli equation with n = 2, and the substitution v = y^{-1} reduces it to the linear equation \frac{dv}{dx} + av = b. The same trick works for any equation of the form

\frac{dy}{dx} = f(x)\,y + g(x)\,y^2

with v = 1/y.

Riccati equations

A Riccati equation has the form

\frac{dy}{dx} = P(x)\,y^2 + Q(x)\,y + R(x)

This is Bernoulli plus an extra R(x) term with no y in it. In general, Riccati equations do not have closed-form solutions — but if you can guess one particular solution y_1(x), then the substitution y = y_1 + \frac{1}{v} reduces the equation to a linear equation in v. You need the lucky guess, but once you have it, the rest is mechanical.

Equations linear in x instead of y

Sometimes an equation is not linear in y but is linear if you swap roles: treat x as the dependent variable and y as the independent variable. That is, work with \frac{dx}{dy} instead of \frac{dy}{dx}. If an equation can be arranged in the form

\frac{dx}{dy} + P(y)\,x = Q(y)

then it is linear in x, and you solve it exactly as before — but with the roles of variables swapped. The integrating factor is e^{\int P(y)\,dy}, and you find x as a function of y.

Example. The equation \frac{dy}{dx} = \frac{1}{y + x} is not linear in y, because the right side has y in a denominator. Flip it: \frac{dx}{dy} = y + x, which rearranges to \frac{dx}{dy} - x = y. This is linear in x. Solve: \mu(y) = e^{-y}, and (e^{-y}x)' = y e^{-y}, so e^{-y}x = -ye^{-y} - e^{-y} + C, giving x = -y - 1 + Ce^y.

Equations of the form y' = f(ax + by + c)

If the right side is a function of a linear combination ax + by + c, try the substitution v = ax + by + c. Then \frac{dv}{dx} = a + b\frac{dy}{dx} = a + b f(v), which is separable.

Example. \frac{dy}{dx} = (x + y + 1)^2. Let v = x + y + 1. Then \frac{dv}{dx} = 1 + \frac{dy}{dx} = 1 + v^2, which separates:

\frac{dv}{1 + v^2} = dx
\arctan v = x + C
v = \tan(x + C)
y = \tan(x + C) - x - 1

One substitution, and a nonlinear equation becomes trivial.

Worked examples

Example 1: Solve dy/dx + y = xy³ (Bernoulli with n = 3)

Step 1. Recognise the form. This is y' + Py = Qy^n with P = 1, Q = x, n = 3.

Why: the right side has a factor of y^3, the coefficient of y on the left is 1, and the coefficient of y^3 is x — all functions of x alone. It is a Bernoulli equation.

Step 2. Substitute v = y^{1-n} = y^{-2}.

Then \frac{dv}{dx} = -2 y^{-3}\frac{dy}{dx}, so y^{-3}\frac{dy}{dx} = -\frac{1}{2}\frac{dv}{dx}.

Why: v = y^{1-n} is the Bernoulli substitution, and n = 3 gives v = y^{-2}. Compute dv/dx by the chain rule.

Step 3. Divide the original equation by y^3 and substitute.

y^{-3}\frac{dy}{dx} + y^{-2} = x
-\frac{1}{2}\frac{dv}{dx} + v = x

Multiply by -2 to put in standard form:

\frac{dv}{dx} - 2v = -2x

Why: the standard form \frac{dv}{dx} + \tilde P(x) v = \tilde Q(x) requires the coefficient of dv/dx to be 1. Multiplying by -2 achieves this and the rest falls into place.

Step 4. Solve the linear equation for v.

The integrating factor is

\mu(x) = e^{\int -2\,dx} = e^{-2x}

Multiply the linear equation by \mu:

e^{-2x}\frac{dv}{dx} - 2e^{-2x}v = -2xe^{-2x}
\frac{d}{dx}\bigl(e^{-2x}v\bigr) = -2xe^{-2x}

Integrate:

e^{-2x}v = \int -2xe^{-2x}\,dx

Integration by parts with u = x, dw = -2e^{-2x}\,dx gives du = dx and w = e^{-2x}:

\int -2xe^{-2x}\,dx = x\,e^{-2x} - \int e^{-2x}\,dx = xe^{-2x} + \frac{1}{2}e^{-2x} + C

So e^{-2x}v = xe^{-2x} + \frac{1}{2}e^{-2x} + C, and

v = x + \frac{1}{2} + Ce^{2x}

Why: the integrating factor method works exactly as before. Integration by parts handles the xe^{-2x} integral, yielding a clean closed form.

Step 5. Back-substitute. Since v = y^{-2},

y^{-2} = x + \frac{1}{2} + Ce^{2x}
y = \pm\frac{1}{\sqrt{x + 1/2 + Ce^{2x}}}

Why: solving y^{-2} = v gives y = \pm v^{-1/2}. The sign depends on the sign of y at the initial condition; it cannot flip during the evolution (the solution would pass through y = 0 at which point y^{-2} blows up).

Result: y = \dfrac{1}{\sqrt{x + 1/2 + Ce^{2x}}} (for y > 0) or its negative.

Two solution curves of $y' + y = xy^3$ with positive $y$. Both curves decrease as $x$ grows because the dominant term $Ce^{2x}$ under the square root grows fast. The solutions are defined on intervals where $x + 1/2 + Ce^{2x} > 0$.

Example 2: The logistic equation, in full

Solve \dfrac{dP}{dt} = rP\left(1 - \dfrac{P}{K}\right) with P(0) = P_0.

Step 1. Rewrite in Bernoulli form.

\frac{dP}{dt} - rP = -\frac{r}{K}P^2

This is Bernoulli with P(t) = -r (coefficient of P), Q(t) = -r/K (coefficient of P^2), and n = 2.

Why: the logistic equation mixes a linear term rP and a quadratic term -rP^2/K. Moving the linear term to the left gives the standard Bernoulli layout with n = 2.

Step 2. Substitute v = P^{1-2} = P^{-1} = 1/P.

Then \frac{dv}{dt} = -P^{-2}\frac{dP}{dt}, so P^{-2}\frac{dP}{dt} = -\frac{dv}{dt}.

Why: for n = 2, the Bernoulli substitution is v = 1/P. This is the one substitution you should memorise if you only memorise one.

Step 3. Divide the Bernoulli equation by P^2 and substitute.

P^{-2}\frac{dP}{dt} - r P^{-1} = -\frac{r}{K}
-\frac{dv}{dt} - rv = -\frac{r}{K}

Multiply by -1:

\frac{dv}{dt} + rv = \frac{r}{K}

Why: the algebra is mechanical once you have the substitution. The result is a linear equation in v with constant coefficients — the simplest kind.

Step 4. Solve the linear equation for v.

Integrating factor: \mu(t) = e^{\int r\,dt} = e^{rt}. Multiply:

e^{rt}\frac{dv}{dt} + re^{rt}v = \frac{r}{K}e^{rt}
\frac{d}{dt}(e^{rt}v) = \frac{r}{K}e^{rt}

Integrate:

e^{rt}v = \frac{1}{K}e^{rt} + C
v = \frac{1}{K} + Ce^{-rt}

Why: one integration, one constant C. The linear equation was as simple as they get.

Step 5. Back-substitute and apply the initial condition.

\frac{1}{P} = \frac{1}{K} + Ce^{-rt}

At t = 0: \frac{1}{P_0} = \frac{1}{K} + C, so C = \frac{1}{P_0} - \frac{1}{K} = \frac{K - P_0}{K P_0}.

Plug back:

\frac{1}{P} = \frac{1}{K} + \frac{K - P_0}{K P_0}e^{-rt}
\frac{1}{P} = \frac{P_0 + (K - P_0)e^{-rt}}{K P_0}
P(t) = \frac{KP_0}{P_0 + (K - P_0)e^{-rt}}

Why: all the algebra is just putting things over a common denominator, then inverting. The final form is the standard logistic curve.

Result: P(t) = \dfrac{KP_0}{P_0 + (K - P_0)e^{-rt}}.

As t \to \infty, the exponential e^{-rt} dies, and P \to K. The population approaches the carrying capacity. As t \to -\infty, the exponential grows without bound, and (for P_0 < K) P \to 0. The logistic curve is S-shaped: slow start, rapid middle, saturation at the top.

The logistic curve for $K = 10$, $P_0 = 1$, $r = 1$: $P(t) = \frac{10}{1 + 9e^{-t}}$. The population starts small, accelerates through the middle region, and saturates at $K = 10$ (dashed red). Every logistic-growth phenomenon — bacterial cultures, species populations, the spread of information — has this shape.

Common confusions

Going deeper

If you can solve Bernoulli equations and spot a few of the other reducible forms, you have the practical toolkit. The rest of this section places these tricks in a wider context — why substitutions work, and where the boundary of what can be "reduced" really sits.

The substitution as a change of variable

A substitution v = g(y) is a change of variable. It replaces the unknown y with a new unknown v, expressed via an invertible function g. The differential equation transforms accordingly: by the chain rule, v' = g'(y)y', so y' = v'/g'(y). Substituting into the original equation gives a new equation for v.

The question "can this equation be reduced to linear form by a substitution" is really "is there a change of variable v = g(y) that transforms the equation into a linear one?" Sometimes the answer is yes (Bernoulli, logistic, the f(ax + by) cases). Sometimes it is no. There is no algorithm that always decides, but there are families — named after Bernoulli, Riccati, d'Alembert, Clairaut — where the right substitution is known.

Riccati equations and the bigger picture

The Riccati equation y' = P(x)y^2 + Q(x)y + R(x) generalises Bernoulli (which is the case R = 0). It is not reducible to a linear equation by any substitution in general — you need to know one particular solution to get anywhere.

But Riccati has a surprising feature. If you set y = -\frac{u'}{P u} for a new function u, the Riccati equation transforms into a linear second-order equation for u:

u'' - \left(Q + \frac{P'}{P}\right)u' + PRu = 0

So Riccati equations can be "reduced" to second-order linear equations. This suggests a deeper principle: the Riccati equation is the first-order shadow of a second-order linear equation, and studying one illuminates the other. This connection is the reason Riccati equations appear in quantum mechanics (the WKB approximation) and in control theory (the algebraic Riccati equation).

Abel equations of the first kind

y' = f_0(x) + f_1(x)y + f_2(x)y^2 + f_3(x)y^3

is called an Abel equation of the first kind. No universal substitution reduces this to linear or separable form; Abel equations are genuinely hard. Some special cases admit closed-form solutions, but in general you have to resort to numerical methods or special functions.

The hierarchy reads: linear < Bernoulli < Riccati < Abel < general. Each step up the chain loses some of the tractability of the previous one. The integrating factor method handles the linear layer completely, Bernoulli adds one more substitution, Riccati adds one guessed solution, and Abel forces you outside elementary methods entirely.

Lie's approach to solvability

The Norwegian mathematician Sophus Lie (pronounced "Lee") developed a general theory of symmetry groups of differential equations. His key observation: an ODE can be solved by quadrature (meaning "integration") if and only if it has a one-parameter group of symmetries. Bernoulli's equation has such a symmetry — y \mapsto \lambda y scales solutions into solutions for an appropriate scaling of \lambda — and that is the reason the substitution v = y^{1-n} works.

The deep reason substitutions work is always a hidden symmetry. Lie's theory gives an algorithm for finding the substitution given the symmetry, and for recognising when no such symmetry exists. It is beautiful mathematics that sits above the mechanical tricks you have just learned — the tricks are what you see when a Lie symmetry is actually present.

Where this leads next

You now have techniques for first-order equations that are linear, Bernoulli, separable, and a few special nonlinear forms. The remaining category of first-order equations that you can solve in closed form is the exact equations — a different approach based on total differentials.