# Bernoulli Differential Equations

• TranscendArcu
In summary, the conversation discusses the possibility of solving a Bernoulli differential equation with n=1 by constructing an integrating factor. The speaker suggests using the equation y' + y[(p + g)(x)] = 0 and multiplying through by Ω(x) to get Ω(x)y = c as a possible solution. The other speaker clarifies that F'(x) = 0 does not imply F(x) = 0, but rather F(x) = constant. This leads to the conclusion that y = \frac{c}{Ω(x)} is a possible solution.
TranscendArcu
Suppose I have a Bernoulli differential equation; that is, an equation of the form: $y' + p(x)y = g(x) y^n$. Supposing that I let $n=1$, the equation is linear. Can I solve it by constructing an integrating factor? That is, can I observe:

$y' + p(x)y = g(x) y$
$→ y' + y[(p + g)(x)] = 0$. I would then have,
$Ω(x) = e^{\int (p + g)(x) dx}$ and multiplying through,
$Ω(x)y' + Ω(x)y[(p + g)(x)] = 0$
$→ (Ω(x)y)' = 0 → Ω(x)y = 0$

But, this seems to be leading me to the conclusion that y = 0. Is that right or have I done something wrong? Is it possible to solve a Bernoulli equation with n=1 by constructing an integrating factor?

TranscendArcu said:
Suppose I have a Bernoulli differential equation; that is, an equation of the form: $y' + p(x)y = g(x) y^n$. Supposing that I let $n=1$, the equation is linear. Can I solve it by constructing an integrating factor? That is, can I observe:

$y' + p(x)y = g(x) y$
$→ y' + y[(p + g)(x)] = 0$. I would then have,
$Ω(x) = e^{\int (p + g)(x) dx}$ and multiplying through,
$Ω(x)y' + Ω(x)y[(p + g)(x)] = 0$
$→ (Ω(x)y)' = 0 → Ω(x)y = 0$

But, this seems to be leading me to the conclusion that y = 0. Is that right or have I done something wrong? Is it possible to solve a Bernoulli equation with n=1 by constructing an integrating factor?

F'(x) = 0 does not imply F(x) = 0; it implies F(x) = constant.

RGV

Oh! So that must mean I have $Ω(x)y = c$. This implies $y = \frac{c}{Ω(x)}$. Not so?

## 1. What is Bernoulli Differential Equation?

Bernoulli Differential Equation is a type of first-order nonlinear ordinary differential equation that can be expressed in the form of dy/dx + P(x)y = Q(x)y^n, where P(x) and Q(x) are continuous functions and n is a constant. It is named after the Swiss mathematician Jacob Bernoulli who first studied these equations in the 18th century.

## 2. What is the difference between Bernoulli and linear differential equations?

The main difference between Bernoulli and linear differential equations is that in Bernoulli equations, the dependent variable y appears with a power other than 1, while in linear equations, y appears with a power of 1. This makes Bernoulli equations more difficult to solve compared to linear equations.

## 3. What are the applications of Bernoulli Differential Equations?

Bernoulli Differential Equations have various applications in physics, engineering, and economics. They are commonly used to model population growth, chemical reactions, and electric circuits. They are also essential in the study of fluid dynamics, specifically in the Bernoulli's principle which states that the total energy of a fluid remains constant along a streamline.

## 4. How do you solve a Bernoulli Differential Equation?

The general method for solving Bernoulli Differential Equations involves transforming the original equation into a linear one by using a substitution such as y = u^(1-n). This will result in a linear equation which can be solved using standard techniques such as separation of variables or integrating factors.

## 5. Are there any special cases of Bernoulli Differential Equations?

Yes, there are two special cases of Bernoulli Differential Equations. The first is when n = 0, which results in a linear equation. The second is when n = 1, which results in a separable equation. These special cases are relatively easier to solve compared to the general Bernoulli equation.

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