In calculus, Jacob Bernoulli introduced the solution for the ordinary differential equation in the following form.

$\dfrac{dy}{dx}+Py$ $\,=\,$ $Qy^{\displaystyle n}$

So, it is called Bernoulli differential equation and let’s learn how to Jacob Bernoulli derived a solution for solving this special first-order and first-degree differential equation.

On right hand side of the equation, the variable $y$ raised to the power of constant $n$ creates some problems while finding the solution of this first order differential equation. So, it is recommendable to remove it from that side in the equation. In order to obtain it, divide both sides of the equation by the variable $y$ raised to the $n$-th power.

$\implies$ $\dfrac{\dfrac{dy}{dx}+Py}{y^{\displaystyle n}}$ $\,=\,$ $\dfrac{Qy^{\displaystyle n}}{y^{\displaystyle n}}$

The division of some of two terms by a function can be calculated by the sum of their quotients.

$\implies$ $\dfrac{\dfrac{dy}{dx}}{y^{\displaystyle n}}+\dfrac{Py}{y^{\displaystyle n}}$ $\,=\,$ $\dfrac{Q \times y^{\displaystyle n}}{y^{\displaystyle n}}$

Now, it is time to simplify the expressions on both sides of the equation.

$\implies$ $\dfrac{\dfrac{dy}{dx} \times 1}{y^{\displaystyle n}}$ $+$ $\dfrac{P \times y}{y^{\displaystyle n}}$ $\,=\,$ $\dfrac{Q \times \cancel{y^{\displaystyle n}}}{\cancel{y^{\displaystyle n}}}$

$\implies$ $\dfrac{dy}{dx} \times \dfrac{1}{y^{\displaystyle n}}$ $+$ $P \times \dfrac{y}{y^{\displaystyle n}}$ $\,=\,$ $\dfrac{Q \times 1}{1}$

On left hand side of the equation, the quotient of the variable $y$ by the $n$-th power of $y$ as per the quotient rule of exponents with same base.

$\implies$ $\dfrac{1}{y^{\displaystyle n}} \times \dfrac{dy}{dx}$ $+$ $P \times y^{1 \displaystyle -n}$ $\,=\,$ $Q$

As per the negative exponential rule, the reciprocal of an exponential function can be written as a negative exponential function.

$\implies$ $y^{\displaystyle -n} \times \dfrac{dy}{dx}$ $+$ $P \times y^{1 \displaystyle -n}$ $\,=\,$ $Q$

$\implies$ $y^{\displaystyle -n} \dfrac{dy}{dx}$ $+$ $P \times y^{1 \displaystyle -n}$ $\,=\,$ $Q$

The simplified ordinary differential equation is very difficult to find its solution. So, it is recommendable to reduce its complexity.

Suppose $z \,=\, y^{1 \displaystyle -n}$, then differentiate this equation with respect to $x$.

$\implies$ $\dfrac{d}{dx}{(z)} \,=\, \dfrac{d}{dx}{\big(y^{1 \displaystyle -n}\big)}$

In this case, $y$ and $z$ are variables, and $n$ is a constant. On left hand side of the equation, the derivative of variable $z$ with respect to $x$ is written as follows. On right hand side of the equation, the derivative of power function is not in terms of $x$ but it should be differentiated with respect to $x$. Hence, the differentiation of the power function can be performed by the combination both power rule of derivatives and chain rule.

$\implies$ $\dfrac{dz}{dx} \,=\, (1-n) \times y^{1 {\displaystyle -n-}1} \times \dfrac{dy}{dx}$

$\implies$ $\dfrac{dz}{dx} \,=\, (1-n) \times y^{\cancel{1} {\displaystyle -n-}\cancel{1}} \times \dfrac{dy}{dx}$

$\implies$ $\dfrac{dz}{dx} \,=\, (1-n) \times y^{\displaystyle -n}\dfrac{dy}{dx}$

$\implies$ $(1-n) \times y^{\displaystyle -n}\dfrac{dy}{dx} \,=\, \dfrac{dz}{dx}$

$\implies$ $y^{\displaystyle -n}\dfrac{dy}{dx} \,=\, \dfrac{1}{1-n} \times \dfrac{dz}{dx}$

$\,\,\,\therefore\,\,\,\,\,\,$ $y^{\displaystyle -n}\dfrac{dy}{dx} \,=\, \dfrac{1}{1-n} \dfrac{dz}{dx}$

The Bernoulli’s differential equation is simplified in the first step as follows.

$y^{\displaystyle -n} \dfrac{dy}{dx}$ $+$ $P \times y^{1 \displaystyle -n}$ $\,=\,$ $Q$

In the above steps, we have derived the following two equations for converting the expressions in terms of $z$ from $y$.

$(1).\,\,\,$ $z \,=\, y^{1 \displaystyle -n}$

$(2).\,\,\,$ $y^{\displaystyle -n}\dfrac{dy}{dx} \,=\, \dfrac{1}{1-n} \dfrac{dz}{dx}$

Now, convert the differential equation in terms of $y$ into $z$ by substituting the above equations.

$\implies$ $\dfrac{1}{1-n} \dfrac{dz}{dx}$ $+$ $P \times z$ $\,=\,$ $Q$

Multiply the both sides of the equation by $1-n$ for eliminating the coefficient of the derivative of $z$ with respect to $x$.

$\implies$ $(1-n) \times \bigg(\dfrac{1}{1-n} \times \dfrac{dz}{dx}$ $+$ $P \times z\bigg)$ $\,=\,$ $(1-n) \times Q$

Now, distribute the factor over the addition of the terms as per the distributive property of multiplication over addition.

$\implies$ $(1-n) \times \dfrac{1}{1-n} \times \dfrac{dz}{dx}$ $+$ $(1-n) \times P \times z$ $\,=\,$ $(1-n) \times Q$

$\implies$ $\dfrac{(1-n) \times 1}{1-n} \times \dfrac{dz}{dx}$ $+$ $(1-n) \times P \times z$ $\,=\,$ $(1-n) \times Q$

$\implies$ $\dfrac{1-n}{1-n} \times \dfrac{dz}{dx}$ $+$ $(1-n) \times P \times z$ $\,=\,$ $(1-n) \times Q$

$\implies$ $\dfrac{\cancel{1-n}}{\cancel{1-n}} \times \dfrac{dz}{dx}$ $+$ $(1-n) \times P \times z$ $\,=\,$ $(1-n) \times Q$

$\implies$ $\dfrac{dz}{dx}$ $+$ $(1-n) \times P \times z$ $\,=\,$ $(1-n) \times Q$

$\implies$ $\dfrac{dz}{dx}$ $+$ $P \times (1-n) \times z$ $\,=\,$ $Q \times (1-n)$

$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{dz}{dx}$ $+$ $P(1-n)z$ $\,=\,$ $Q(1-n)$

The first order and first degree differential equation in complex form is simplified to Leibnitz or Leibniz’s linear differential equation.

Latest Math Topics

Jan 06, 2023

Jan 03, 2023

Jan 01, 2023

Dec 26, 2022

Dec 08, 2022

Latest Math Problems

Nov 25, 2022

Nov 02, 2022

Oct 26, 2022

Oct 24, 2022

Sep 30, 2022

A best free mathematics education website for students, teachers and researchers.

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Learn how to solve the maths problems in different methods with understandable steps.

Copyright © 2012 - 2022 Math Doubts, All Rights Reserved