# Meeting Mr. Bernoulli The differential equation ${d \over dx} y + f(x) y = g(x) y^{\alpha}\quad\quad\quad(1)$

where $\alpha \neq 0, 1$ and $g(x) \not \equiv 0$, is known as the Bernoulli’s equation.

When $\alpha$ is an integer, (1) has trivial solution $y(x) \equiv 0$.

To obtain nontrivial solution, we divide each term of (1) by $y^{\alpha}$ to get, $\boxed{y^{-\alpha}{d \over dx}y} + f(x) y^{1-\alpha} = g(x)\quad\quad\quad(2)$

Since ${d \over {dx}}({{1 \over {1-\alpha}}y^{1-\alpha}}) ={1 \over {1-\alpha}}\cdot (1-\alpha) y^{1-\alpha-1}{d \over dx}y=\boxed{y^{-\alpha}{d \over dx}y}$

(2) can be expressed as ${d \over dx} ({{1 \over {1-\alpha}} y^{1-\alpha}}) + f(x) y^{1-\alpha} = g(x)$

which is ${{1 \over {1-\alpha}} {d \over dx} y^{1-\alpha}} + f(x) y^{1-\alpha} = g(x)$ .

Multiply $1-\alpha$ throughout, ${d \over dx} y^{1-\alpha} + (1-\alpha) f(x) y^{1-\alpha} = (1-\alpha) g(x)\quad\quad\quad(3)$

Let $z = y^{1-\alpha}$, (3) is transformed to a first order linear equation ${d \over dx} z + (1-\alpha) f(x) z = (1-\alpha) g(x)$,

giving the general solution of a Bernoulli’s equation (see Fig. 1) Fig. 1

For a concrete example of Bernoulli’s equation, see “What moves fast, will slow down