Definition of Bernoulli ODEs
Bernoulli ODEs are first-order ODEs in the form of
\begin{equation}
\frac{dy}{dx} + P(x)y = Q(x)y^n \tag{1}
\end{equation}
where \( P(x), Q(x) \) are functions of \( x \), and \( n \neq 0,1 \). This means that Bernoulli ODEs are a kind of modified first-order linear ODEs (see this tutorial) where the source term \( Q(x) \) is now combined with some power of \( y \). Notice that if \( n = 0 \) or \( 1 \) then it just reduces to a simple first-order linear ODE. Some examples of Bernoulli ODEs are
- \( dy/dx + 3e^{x}y = x^2y^3\);
- \( dy/dx + 5y = y^4 \).
Substitution Method
Bernoulli ODEs can be solved by the substitution \( v = y^{1-n}\) that transforms it into a linear ODE. Note that
\begin{align}
\frac{dv}{dx} &= (1-n)y^{-n}\frac{dy}{dx} \\
\frac{dy}{dx} &= \frac{y^n}{1-n} \frac{dv}{dx} \tag{2}
\end{align}
Hence plugging it into (1) gives
\begin{align}
\frac{y^n}{1-n} \frac{dv}{dx} + P(x)y &= Q(x)y^n \\
\frac{1}{1-n} \frac{dv}{dx} + P(x)y^{1-n} &= Q(x) \\
\frac{1}{1-n} \frac{dv}{dx} + P(x)v &= Q(x) \tag{3}
\end{align}
which can be tackled using the integrating factor method introduced previously.
Example
We will approach the Logistic Population Model problem again, but now from the perspective of Bernoulli ODEs:
\begin{equation}
\frac{dN}{dt} = KN(M-N) = KMN -KN^2 \tag{4}
\end{equation}
\( n = 2 \) in this case. So we can make a substitution of \( V = N^{1-2} = N^{-1} \), and (2) looks like
\begin{align}
\frac{dN}{dt} &= -N^2 \frac{dV}{dt} \tag{5}
\end{align}
Subsequently, (4) becomes
\begin{align}
-N^2 \frac{dV}{dt} &= KMN -KN^2 \\
\frac{dV}{dt} &= K -\frac{KM}{N} = K -KMV \tag{6}
\end{align}
This is indeed a first-order ODE with the integrating factor of \( e^{KMt} \):
\begin{align}
\frac{dV}{dt} + KMV &= K \\
e^{KMt}\frac{dV}{dt} + KMVe^{KMt} &= Ke^{KMt} \\
\frac{d}{dt}(e^{KMt}V) &= Ke^{KMt} \\
e^{KMt}V &= \int Ke^{KMt} dt \\
e^{KMt}V &= \frac{1}{M}e^{KMt} + C \\
V &= \frac{1}{M} + Ce^{-KMt} \tag{7}
\end{align}
Recovering the original variable then yields the same solution as before:
\begin{align}
\frac{1}{N} &= \frac{CMe^{-KMt} + 1}{M} \\
N &= \frac{M}{CMe^{-KMt} + 1} = \frac{Me^{KMt}}{CM + e^{KMt}} \\
&= \frac{MAe^{KMt}}{1 + Ae^{KMt}} \tag{8}
\end{align}
where we set \( A = 1/CM \) here.
Exercise
Show that the following is a Bernoulli ODE and solve it:
\begin{equation}
\frac{dy}{dx} + xy = x^3y^3 \tag{9}
\end{equation}
Answer
Here \( n = 3\). Let \( v = y^{1-3} = y^{-2} = 1/y^2 \), \( y = v^{-1/2} = 1/\sqrt{v} \), then
\begin{align}
\frac{dy}{dx} &= -\frac{1}{2}v^{-3/2} \frac{dv}{dx} = -\frac{y^3}{2} \frac{dv}{dx}
\end{align}
and the ODE becomes
\begin{align}
-\frac{y^3}{2} \frac{dv}{dx} + xy &= x^3y^3 \\
-\frac{1}{2} \frac{dv}{dx} + xy^{-2} &= x^3 \\
\frac{dv}{dx} -2xv &= -2x^3
\end{align}
The integrating factor is \( \exp(\int -2xdx) = \exp(-x^2) \):
\begin{align}
e^{-x^2}\frac{dv}{dx} -2xe^{-x^2}v &= -2x^3e^{-x^2} \\
\frac{d}{dx}(e^{-x^2}v) &= -2x^3e^{-x^2} \\
e^{-x^2}v &= -2 \int x^3e^{-x^2} dx \\
e^{-x^2}v &= -\int x^2e^{-x^2} d(x^2) \\
e^{-x^2}v &= \int x^2 d(e^{-x^2}) \\
e^{-x^2}v &= [x^2 e^{-x^2}] -\int e^{-x^2} d(x^2) \\
e^{-x^2}v &= x^2 e^{-x^2} + e^{-x^2} + C \\
v &= x^2 + 1 + Ce^{x^2} \\
y &= \frac{1}{\sqrt{x^2 + 1 + Ce^{x^2}}}
\end{align}
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