Bernoulli Differential Equations

# Bernoulli Differential Equations

We are now going to look at a method for solving another class of differential equations. Let $p(x)$ and $g(x)$ be continuous on an interval of interest, and consider the following non-linear differential equation:

(1)
\begin{align} \quad y' + p(x)y = g(x)y^n \end{align}

If either $n = 0$ or $n = 1$, the differential equation above is linear, so suppose that $n > 1$. We can "convert" this differential equation to be in a linear form using some substitutions. Let $v = y^{1-n}$. Then $v' = (1 - n)y^{-n}y'$. Now take the differential equation above and divide both sides by $y^n$ to get that:

(2)
\begin{align} \quad y^{-n} y' + p(x) y^{1-n} = g(x) \end{align}

We can now apply the substitutions we made above to get that:

(3)
\begin{align} \quad \left ( \frac{1}{1 - n} \right ) v' + p(x) v = g(x) \end{align}

This differential equation is linear, and we can solve this differential equation using the method of integrating factors. The important thing to remember for Bernoulli Differential Equations is that we make the following substitutions:

(4)

Let's look at some examples of solving Bernoulli Differential equations.

## Example 1

Solve the differential equation $y' + 2xy = 3xy^3$.

We first start off by dividing the differential equation above by $y^3$ to get:

(5)
\begin{align} \quad y^{-3} y' + 2x y^{-2} = 3x \end{align}

Now we will make the substitution $v = y^{1-3} = y^{-2}$ and so $v' = -2 y^{-3} y'$. so $y^{-3}y' = 1\frac{1}{2} v'$. Applying these substitutions and we get:

(6)
\begin{align} \quad -\frac{1}{2} v' + 2x v = 3x \\ \quad v' - 4xv = -6x \end{align}

Let's solve this differential equation using integrating factors. Let $\mu (t) = e^{\int -4x \: dx} = e^{-2x^2}$. Multiplying both sides of the differential equation above by the integrating factor and we have that:

(7)
\begin{align} \quad \mu (t) v' - \mu (t) 4x v = -\mu(t) 6x \\ \quad e^{-2x^2} v' - e^{-2x^2} 4x v = -6x e^{-2x^2} \\ \quad \frac{d}{dt} \left ( e^{-2x^2} v \right ) = -6x e^{-2x^2} \\ \quad e^{-2x^2}v = \int -6x e^{-2x^2} \: dx \end{align}

We can solve the integral on the right by using substitution. Let $u = -2x^2$. Then $du = -4x \: dx$ and so $\frac{3}{2} du = -6x \: dx$. Hence:

(8)
\begin{align} \quad e^{-2x^2}v = \frac{3}{2} \int e^{u} \: dx \\ \quad e^{-2x^2}v = \frac{3}{2} e^{u} + C \\ \quad e^{-2x^2}v = \frac{3}{2} e^{-2x^2} + C \\ \quad v = \frac{3}{2} + \frac{C}{e^{-2x^2}} \end{align}

We are almost done. We will now use the substitution $v = y^{-2}$ that we started with to get that:

(9)
\begin{align} \quad y^{-2} = \frac{3}{2} + \frac{C}{e^{-2x^2}} \\ \quad y^2 = \frac{1}{\frac{3}{2} + \frac{C}{e^{-2x^2}}} \end{align}