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

May 21, 2023

May 16, 2023

May 10, 2023

May 03, 2023

Latest Math Problems

May 09, 2023

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

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

A math help place with list of solved problems with answers and worksheets on every concept for your practice.

Copyright © 2012 - 2022 Math Doubts, All Rights Reserved