# 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)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)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)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)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)Therefore we have that:

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

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

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

(11)