Bernoulli Equation

Bernoulli Equation

\[y'+p(x)y=g(x)y^a\quad \text{(}a\text{ is any real number)} \tag{1}\]

The Bernoulli equation (1) is linear if $a=0$ or $a=1$, and nonlinear otherwise. However, it can be transformed into a linear equation through the following process.

Let \(u(x)=[y(x)]^{1-a}\)

Differentiating this and substituting $y’$ from equation (1), we get

\[\begin{align*} u'&=(1-a)y^{-a}y' \\&=(1-a)y^{-a}(gy^a-py) \\&=(1-a)(g-py^{1-a}) \end{align*}\]

In the right-hand side, $y^{1-a}=u$, so we obtain the following linear first-order differential equation:

\[u'+(1-a)pu=(1-a)g \tag{2}\]

Example: Logistic Equation

Solve the logistic equation (a special form of the Bernoulli equation).

\[y'=Ay-By^2 \tag{3}\]

Solution

Writing equation (3) in the form of equation (1):

\[y'-Ay=-By^2\]

Here, $a=2$, so $u=y^{1-a}=y^{-1}$. Differentiating this u and substituting $y’$ from equation (3):

\[u'=-y^{-2}y'=-y^{-2}(Ay-By^2)=B-Ay^{-1}\]

The last term is $-Ay^{-1}=-Au$, so we obtain the following linear first-order differential equation:

\[u'+Au=B\]

By the solution formula for non-homogeneous linear first-order differential equations, we can find the following general solution:

\[u=ce^{-At}+B/A\]

Since $u=1/y$, we obtain the general solution of equation (3):

\[y=\frac{1}{u}=\frac{1}{ce^{-At}+B/A} \tag{4}\]