Many nonlinear equations can be turned into ones we already know how to solve by a change of variables. The two classic families are homogeneous equations, which become separable, and Bernoulli equations, which become linear.
Key idea
Homogeneous: if $y' = F\!\left(\frac{y}{x}\right)$, set $v = \frac{y}{x}$, so $y = vx$ and $y' = v + xv'$. Then $xv' = F(v) - v$ is separable.
Bernoulli: if $y' + p(t)y = q(t)y^{n}$ with $n \ne 0, 1$, set $v = y^{1-n}$. Then $v' + (1 - n)p\,v = (1 - n)q$ is linear.
Bernoulli: if $y' + p(t)y = q(t)y^{n}$ with $n \ne 0, 1$, set $v = y^{1-n}$. Then $v' + (1 - n)p\,v = (1 - n)q$ is linear.
Homogeneous equations
Why it works
A function of $y/x$ alone does not change if $x$ and $y$ are scaled together, so after writing $y = vx$ the right side depends only on $v$. The only place $x$ survives is the product rule term $xv'$, and dividing by $x$ separates the variables: $\frac{dv}{F(v) - v} = \frac{dx}{x}$.
Worked example: A homogeneous equation
Solve $xy\,y' = x^{2} + y^{2}$, $y(1) = 2$, for $x > 0$.
- Divide by $xy$: $y' = \frac{x}{y} + \frac{y}{x}$, a function of $v = \frac{y}{x}$: $F(v) = \frac{1}{v} + v$.
- With $y' = v + xv'$: $v + xv' = \frac{1}{v} + v$, so $xv' = \frac{1}{v}$.
- Separate: $v\,dv = \frac{dx}{x}$, so $\frac{v^{2}}{2} = \ln x + C$.
- At $x = 1$, $v = 2$: $C = 2$, so $v^{2} = 2\ln x + 4$.
Answer: $y = x\sqrt{2\ln x + 4}$
Bernoulli equations
Why it works
Divide $y' + py = qy^{n}$ by $y^{n}$: $$y^{-n}y' + p\,y^{1-n} = q.$$ The chain rule gives $(y^{1-n})' = (1 - n)y^{-n}y'$, so with $v = y^{1-n}$ the first term is $\frac{v'}{1 - n}$ and the second is $pv$. Multiplying by $1 - n$ leaves a linear equation for $v$.
Worked example: A Bernoulli equation
Solve $y' + y = y^{2}$, $y(0) = \frac{1}{2}$.
- $n = 2$, so $v = y^{-1}$ and $v' = -y^{-2}y'$.
- Divide by $y^{2}$: $y^{-2}y' + y^{-1} = 1$, which is $-v' + v = 1$, or $v' - v = -1$.
- Linear: $\mu = e^{-t}$, $(e^{-t}v)' = -e^{-t}$, so $v = 1 + Ce^{t}$.
- $v(0) = \frac{1}{y(0)} = 2$ gives $C = 1$.
Answer: $y = \frac{1}{1 + e^{t}}$
Worked example: The logistic equation as a Bernoulli equation
Solve $P' = rP - \frac{r}{K}P^{2}$.
- Rewrite as $P' - rP = -\frac{r}{K}P^{2}$: Bernoulli with $n = 2$. Let $v = \frac{1}{P}$.
- Then $v' + rv = \frac{r}{K}$, whose solution is $v = \frac{1}{K} + Ae^{-rt}$.
- So $P = \frac{1}{\frac{1}{K} + Ae^{-rt}} = \frac{K}{1 + AKe^{-rt}}$.
Answer: The familiar logistic curve from The Logistic Equation, found this time without partial fractions.
Common mistake
Common slips: substituting $v = y^{n}$ instead of $y^{1-n}$; substituting before dividing by $y^{n}$; and forgetting the factor $1 - n$ in $v'$. Also, dividing by $y^{n}$ hides the solution $y = 0$ (when $n > 0$). Check at the end by substituting your $y$ back into the original equation.