Bernoulli Differential Equations Examples 1

# Bernoulli Differential Equations Examples 1

Recall from the Bernoulli Differential Equations page that a differential equation in the form $y' + p(x) y = g(x) y^n$ is called a Bernoulli differential equation. These differential equations are not linear, however, we can "convert" them to be linear.

We first let $v = y^{1-n}$. Then $v' = (1 - n)y^{-n}y'$. We then take the differential equation above and divide both sides of it by $y^n$ and apply these substitutions to get:

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

We then solve this differential equation as a first order linear equation for $v$, and subsequently solve $y$ from $v$.

Let's look at a few examples of solving Bernoulli differential equations.

## Example 1

Solve the differential equation $6y' -2y = ty^4$.

It's not hard to see that this is indeed a Bernoulli differential equation. We first divide by $6$ to get this differential equation in the appropriate form:

(2)
\begin{align} \quad y' - \frac{1}{3} y = \frac{t}{6} y^4 \end{align}

In this case, we have that $n = 4$. Let $v = y^{1 - 4} = y^{-3}$. Then $v' = -3y^{-4}y'$ and $- \frac{v'}{3} = y^{-4}y'$. Now we divide both sides of the differential equation above by $y^4$ and apply these substitutions to get:

(3)
\begin{align} \quad y' - \frac{1}{3} y = \frac{t}{6} y^4 \\ \quad y^{-4}y' - \frac{1}{3} y^{-3} = \frac{t}{6} \\ \quad -\frac{1}{3}v' - \frac{1}{3} v = \frac{t}{6} \\ \quad 2v' + 2v = -t \\ \quad v' + v = -\frac{t}{2} \end{align}

Let's solve this differential equation with integrating factors. We have that $p(t) = 1$, so $\mu(t) = e^{\int p(t) \: dt} = e^{\int 1 \: dt} = e^t$. We multiply both sides of the differential equation above by our integrating factor and we have that:

(4)
\begin{align} \quad e^t v' + e^t v = -\frac{1}{2}t e^t \\ \quad \frac{d}{dt} (e^t v) = -\frac{1}{2}t e^t \\ \quad e^t v = \int -\frac{1}{2}t e^t \: dt \\ \end{align}

We can now use integration by parts to evaluate the integral on the righthand side. Let $u = -\frac{1}{2}t$ and $dv = e^t \: dt$. Then $du = -\frac{1}{2} \: dt$ and $v = e^t$. Therefore:

(5)
\begin{align} \quad \int u \: dv = uv - \int v \: du \\ \quad \int -te^t \: dt = -\frac{1}{2}te^t + \frac{1}{2} \int e^t \: dt \\ \quad \int -te^t \: dt = -\frac{1}{2}te^t + \frac{1}{2}e^t \\ \quad \int -te^t \: dt = \frac{1}{2}(1 - t)e^t + C \end{align}

Therefore we have that:

(6)
\begin{align} \quad e^t v = \frac{1-t}{2}e^t + C \\ \quad v = \frac{1-t}{2} + Ce^{-t} \\ \end{align}

Now we made the substitution that $v = y^{-3}$ earlier. Therefore:

(7)
\begin{align} \quad \frac{1}{y^3} = \frac{1-t}{2} + Ce^{-t} \\ \quad y^3 = \frac{1}{\frac{1 - t}{2} + Ce^{-t}} \end{align}

## Example 2

Solve the differential equation $y' - 5y = - 5ty^3$.

Once again, the differential equation above is indeed a Bernoulli equation. It's already in the appropriate form and $n = 3$, so let $v = y^{1 -n} = y^{-2}$. Then $v' = -2y^{-3} y'$. We take the differential equation above and divide both sides by $y^3$ and apply our substitutions to get:

(8)
\begin{align} \quad y^{-3} y' - 5y^{-2} = -5t \\ \quad -\frac{1}{2} v' -5v = -5t \\ \quad v' + 10v = 10t \end{align}

Let's solve this differential equation using integrating factors again. We have that $p(t) = 10$. Therefore $\mu (t) = e^{10t}$. Multiplying both sides of our differential equation above by this gives us:

(9)
\begin{align} \quad e^{10t}v' + 10e^{10t}v = 10te^{10t} \\ \quad \frac{d}{dt} (e^{10t} v) = 10te^{10t} \\ \quad e^{10t}v = 10 \int t e^{10t} \: dt \\ \end{align}

Let $u = t$ and $dv = e^{10t} \: dt$. Then $du = \: dt$ and $v = \frac{1}{10} e^{10t}$ so:

(10)
\begin{align} \quad e^{10t}v = 10 \left ( \frac{t}{10} e^{10t} - \frac{1}{10} \int e^{10t} \: dt \right ) \\ \quad e^{10t}v = te^{10t} - \frac{1}{10}e^{10t} + C \\ \quad v = t - \frac{1}{10} + Ce^{-10t} \end{align}

Now we have that $v = y^{-2}$ and so:

(11)
\begin{align} \quad y^{-2} = t - \frac{1}{10} + Ce^{-10t} \\ \quad y^2 = \frac{1}{t - \frac{1}{10} + Ce^{-10t}} \\ \end{align}