# Bernoulli’s Equation And Solution

Differential Equation / Monday, September 10th, 2018

Bernoulli Equation is the first order differential equation which can be reducible to linear differential equation

# Definition of Bernoulli’s Equation

An equation of the form
$\frac{dy}{dx}+Py=Q{{y}^{n}}$
where P and Q are functions of x alone, or constants, and n is any rational number, is known as Bernoulli’s Equation.
This equation can be easily reduced to the standard linear form by proper change of variable.
We multiply both sides of the equation by (1 – n)y-n to get
$\left( 1-n \right){{y}^{-n}}\frac{dy}{dx}+\left( 1-n \right)P.{{y}^{1-n}}=\left( 1-n \right)Q$
Now we substitute
${{y}^{1-n}}=z\Rightarrow \left( 1-n \right){{y}^{-n}}\frac{dy}{dx}=\frac{dz}{dx}$
So the equation reducible to
$\frac{dz}{dx}+\left( 1-n \right)P.z=\left( 1-n \right)Q$
This being a linear differential equation in z can be solved by the method that we discussed in the article.

 Example 01

$Solve:\frac{dy}{dx}+\frac{y}{x}={{y}^{2}}$
Solution:
$Given,\frac{dy}{dx}+\frac{y}{x}={{y}^{2}}$
$\Rightarrow \frac{1}{{{y}^{2}}}\frac{dy}{dx}+\frac{1}{y}.\frac{1}{x}=1$
$Putting,\frac{1}{y}=z\Rightarrow -\frac{1}{{{y}^{2}}}\frac{dy}{dx}=\frac{dz}{dx}$
Therefore the given equation becomes
$-\frac{dz}{dx}+z.\frac{1}{x}=1$
$\Rightarrow \frac{dz}{dx}-z.\frac{1}{x}=-1……….(1)$
It is a linear differential equation.
$\therefore I.F.={{e}^{-\int{\frac{1}{x}dx}}}={{e}^{-\log x}}={{e}^{\log {{x}^{-1}}}}=\frac{1}{x}$
Multiplying both sides of equation (1) by I.F. and integrating, we get
$z.\frac{1}{x}=-\int{\frac{1}{x}dx+c}$
$\Rightarrow \frac{z}{x}=-\log x+c$
$\therefore \frac{1}{xy}=-\log x+c$
is the required solution.

 Example 02

$Solve:\frac{dy}{dx}+x\sin 2y={{x}^{3}}{{\cos }^{2}}y$
Solution:
$Given,\frac{dy}{dx}+x\sin 2y={{x}^{3}}{{\cos }^{2}}y$
$\Rightarrow \frac{1}{{{\cos }^{2}}y}\frac{dy}{dx}+\frac{x\sin 2y}{{{\cos }^{2}}y}={{x}^{3}}$
$\Rightarrow {{\sec }^{2}}y\frac{dy}{dx}+\frac{2x\sin y\cos y}{{{\cos }^{2}}y}={{x}^{3}}$
$\Rightarrow {{\sec }^{2}}y\frac{dy}{dx}+2x\tan y={{x}^{3}}$
$Putting,\tan y=z\Rightarrow {{\sec }^{2}}y\frac{dy}{dx}=\frac{dz}{dx}$
Therefore the given equation becomes
$\frac{dz}{dx}+2xz={{x}^{3}}………(1)$
It is a linear differential equation
$\therefore I.F.={{e}^{\int{2xdx}}}={{e}^{{{x}^{2}}}}$
Multiplying both sides of equation (1) by I.F. and integrating, we get
$z{{e}^{{{x}^{2}}}}=\int{{{x}^{3}}{{e}^{{{x}^{2}}}}}dx+c$
$\Rightarrow z{{e}^{{{x}^{2}}}}=\int{{{x}^{2}}.}{{e}^{{{x}^{2}}}}.xdx+c$
$Putting,{{x}^{2}}=k\Rightarrow xdx=\frac{1}{2}dk$
$\therefore z{{e}^{{{x}^{2}}}}=\int{k{{e}^{k}}.}\frac{1}{2}dk+c$
$\Rightarrow z{{e}^{{{x}^{2}}}}=\frac{1}{2}\left[ k\int{{{e}^{k}}dk}-\int{\left\{ \frac{d}{dk}k\int{{{e}^{k}}dk} \right\}dk} \right]+c$
$\Rightarrow z{{e}^{{{x}^{2}}}}=\frac{1}{2}\left[ k{{e}^{k}}-\int{{{e}^{k}}dk} \right]+c$
$\Rightarrow z{{e}^{{{x}^{2}}}}=\frac{1}{2}\left[ k{{e}^{k}}-{{e}^{k}} \right]+c$
$\Rightarrow z{{e}^{{{x}^{2}}}}=\frac{1}{2}{{e}^{{{x}^{2}}}}\left( {{x}^{2}}-1 \right)+c$
$\therefore {{e}^{{{x}^{2}}}}\tan y=\frac{1}{2}{{e}^{{{x}^{2}}}}\left( {{x}^{2}}-1 \right)+c$
is the required solution.

 Example 03

$Solve:\frac{dy}{dx}+\frac{1}{x}=\frac{{{e}^{y}}}{{{x}^{2}}}$
Solution:
$Given,\frac{dy}{dx}+\frac{1}{x}=\frac{{{e}^{y}}}{{{x}^{2}}}$
$\Rightarrow {{e}^{-y}}\frac{dy}{dx}+\frac{{{e}^{-y}}}{x}=\frac{1}{{{x}^{2}}}$
$Putting,{{e}^{-y}}=z\Rightarrow -{{e}^{-y}}\frac{dy}{dx}=\frac{dz}{dx}$
Therefore the given equation becomes
$-\frac{dz}{dx}+\frac{z}{x}=\frac{1}{{{x}^{2}}}$
$\frac{dz}{dx}-\frac{z}{x}=-\frac{1}{{{x}^{2}}}……….(1)$
It is a linear differential equation.
$\therefore I.F.={{e}^{-\int{\frac{1}{x}dx}}}={{e}^{-\log x}}={{e}^{\log {{x}^{-1}}}}=\frac{1}{x}$
Multiplying both sides of equation (1) by I.F. and integrating, we get
$z.\frac{1}{x}=-\int{\frac{1}{x}}.\frac{1}{{{x}^{2}}}dx+c$
$\Rightarrow \frac{z}{x}=\frac{1}{2{{x}^{2}}}+c$
$\therefore \frac{{{e}^{-y}}}{x}=\frac{1}{2{{x}^{2}}}+c$
is the required solution.

 Example 04

$Solve:\frac{dy}{dx}+y\cot x=2{{y}^{2}}\cos x$
Solution:
$Given,\frac{dy}{dx}+y\cot x=2{{y}^{2}}\cos x$
$\Rightarrow \frac{1}{{{y}^{2}}}\frac{dy}{dx}+\frac{1}{y}\cot x=2\cos x$
$Putting,\frac{1}{y}=z\Rightarrow -\frac{1}{{{y}^{2}}}\frac{dy}{dx}=\frac{dz}{dx}$
Therefore the given equation becomes
$-\frac{dz}{dx}+z\cot x=2\cos x$
$\frac{dz}{dx}-z\cot x=-2\cos x……….(1)$
It is a linear differential equation.
$\therefore I.F.={{e}^{-\int{\cot xdx}}}={{e}^{-\log \sin x}}={{e}^{\log \sin {{x}^{-1}}}}=\frac{1}{\sin x}$
Multiplying both sides of equation (1) by I.F. and integrating, we get
$z.\frac{1}{\sin x}=-2\int{\frac{1}{\sin x}.\cos xdx+c}$
$\Rightarrow \frac{z}{\sin x}=-2\log \sin x+c$
$\therefore \frac{1}{y\sin x}=-2\log \sin x+c$
is the required solution.

 Example 05

$Solve:\frac{dy}{dx}+\frac{y\log y}{x}=\frac{y}{{{x}^{2}}}{{\left( \log y \right)}^{2}}$
Solution:
$Given,\frac{dy}{dx}+\frac{y\log y}{x}=\frac{y}{{{x}^{2}}}{{\left( \log y \right)}^{2}}$
Dividing the given equation by y(log y)2, the equation becomes
$\frac{1}{y{{\left( \log y \right)}^{2}}}\frac{dy}{dx}+\frac{1}{x\log y}=\frac{1}{{{x}^{2}}}$
$Putting,\frac{1}{\log y}=z\Rightarrow -\frac{1}{y{{\left( \log y \right)}^{2}}}\frac{dy}{dx}=\frac{dz}{dx}$
Therefore the given equation becomes
$-\frac{dz}{dx}+\frac{1}{x}.z=\frac{1}{{{x}^{2}}}$
$\Rightarrow \frac{dz}{dx}-\frac{1}{x}.z=-\frac{1}{{{x}^{2}}}……….(1)$
It is a linear differential equation.
$\therefore I.F.={{e}^{-\int{\frac{1}{x}dx}}}={{e}^{-\log x}}={{e}^{\log {{x}^{-1}}}}=\frac{1}{x}$
Multiplying both sides of equation (1) by I.F. and integrating, we get
$z.\frac{1}{x}=-\int{\frac{1}{x}}.\frac{1}{{{x}^{2}}}dx+c$
$\Rightarrow \frac{z}{x}=\frac{1}{2{{x}^{2}}}+c$
$\therefore \frac{1}{x\log y}=\frac{1}{2{{x}^{2}}}+c$
is the required solution.

 Example 06

$Solve:x\frac{dy}{dx}+\sin 2y={{x}^{4}}{{\cos }^{2}}y$
Solution:
$Given,x\frac{dy}{dx}+\sin 2y={{x}^{4}}{{\cos }^{2}}y$
$\Rightarrow {{\sec }^{2}}y\frac{dy}{dx}+\frac{2\sin y\cos y}{x{{\cos }^{2}}y}={{x}^{3}}$
$\Rightarrow {{\sec }^{2}}y\frac{dy}{dx}+\frac{2}{x}\tan y={{x}^{3}}$

$Putting,\tan y=z\Rightarrow {{\sec }^{2}}y\frac{dy}{dx}=\frac{dz}{dx}$
Therefore the given equation becomes
$\frac{dz}{dx}+\frac{2}{x}.z={{x}^{3}}……….(1)$
It is a linear differential equation in z.
$\therefore I.F.={{e}^{\int{\frac{2}{x}dx}}}={{e}^{2\log x}}={{e}^{\log {{x}^{2}}}}={{x}^{2}}$
Multiplying the equation (1) by I.F. and integrating, we get
$z.{{x}^{2}}=\int{{{x}^{3}}.{{x}^{2}}}dx+c$
$\Rightarrow z{{x}^{2}}=\frac{{{x}^{6}}}{6}+c\Rightarrow 6z{{x}^{6}}={{x}^{6}}+c$
$\therefore 6{{x}^{6}}\tan y={{x}^{6}}+c$
is the required solution.

 Example 07

$Solve:\frac{dy}{dx}-\frac{\tan y}{1+x}=\left( 1+x \right){{e}^{x}}\sec y$
Solution:
$Given,\frac{dy}{dx}-\frac{\tan y}{1+x}=\left( 1+x \right){{e}^{x}}\sec y$
$\Rightarrow \cos y\frac{dy}{dx}-\cos y.\frac{\tan y}{1+x}=\left( 1+x \right){{e}^{x}}$
$\Rightarrow \cos y\frac{dy}{dx}-\frac{\sin y}{1+x}=\left( 1+x \right){{e}^{x}}$
$Putting,\sin y=z\Rightarrow \cos y\frac{dy}{dx}=\frac{dz}{dx}$
Therefore the given equation becomes
$\frac{dz}{dx}-\frac{z}{1+x}=\left( 1+x \right){{e}^{x}}……….(1)$
It is a linear differential equation in z.
$\therefore I.F.={{e}^{-\int{\frac{1}{1+x}dx}}}={{e}^{-\log \left( 1+x \right)}}={{e}^{\log {{\left( 1+x \right)}^{-1}}}}=\frac{1}{1+x}$
Multiplying the equation (1) by I.F. and integrating, we get
$z.\frac{1}{1+x}=\int{\frac{1}{1+x}}.\left( 1+x \right){{e}^{x}}dx+c$
$\Rightarrow \frac{z}{1+x}={{e}^{x}}+c$
$\therefore \frac{\sin y}{1+x}={{e}^{x}}+c$
is the required solution.

 Example 08

$Solve:\frac{dy}{dx}+\frac{x}{1-{{x}^{2}}}y=x\sqrt{y}$
Solution:
$Given,\frac{dy}{dx}+\frac{x}{1-{{x}^{2}}}y=x\sqrt{y}$
$\Rightarrow {{y}^{-\frac{1}{2}}}\frac{dy}{dx}+\frac{x}{1-{{x}^{2}}}{{y}^{\frac{1}{2}}}=x$
$Putting,{{y}^{\frac{1}{2}}}=z\Rightarrow \frac{1}{2}{{y}^{-\frac{1}{2}}}\frac{dy}{dx}=\frac{dz}{dx}$
Therefore the given equation becomes
$2\frac{dz}{dx}+\frac{x}{1-{{x}^{2}}}z=x$
$\Rightarrow \frac{dz}{dx}+\frac{1}{2}.\frac{x}{1-{{x}^{2}}}z=\frac{1}{2}x……….(1)$
It is a linear differential equation in z.
$\therefore I.F.={{e}^{\frac{1}{2}\int{\frac{x}{1-{{x}^{2}}}dx}}}={{e}^{-\frac{1}{4}\log \left( 1-{{x}^{2}} \right)}}={{e}^{\log {{\left( 1-{{x}^{2}} \right)}^{-\frac{1}{4}}}}}={{\left( 1-{{x}^{2}} \right)}^{-\frac{1}{4}}}$
Multiplying the equation (1) by I.F. and integrating, we get
$z.{{\left( 1-{{x}^{2}} \right)}^{-\frac{1}{4}}}=\frac{1}{2}\int{x.{{\left( 1-{{x}^{2}} \right)}^{-\frac{1}{4}}}dx+c}$
$\Rightarrow z.{{\left( 1-{{x}^{2}} \right)}^{-\frac{1}{4}}}=-\frac{1}{4}\int{\left( -2x \right).{{\left( 1-{{x}^{2}} \right)}^{-\frac{1}{4}}}dx+c}$
$Putting,1-{{x}^{2}}=k\Rightarrow -2xdx=dk$
$\therefore z.{{\left( 1-{{x}^{2}} \right)}^{-\frac{1}{4}}}=-\frac{1}{4}\int{{{k}^{-\frac{1}{4}}}}dk+c$
$\Rightarrow z.{{\left( 1-{{x}^{2}} \right)}^{-\frac{1}{4}}}=-\frac{1}{4}.\frac{4}{3}{{k}^{\frac{3}{4}}}+c$
$\therefore {{y}^{\frac{1}{2}}}{{\left( 1-{{x}^{2}} \right)}^{-\frac{1}{4}}}=-\frac{1}{3}{{\left( 1-{{x}^{2}} \right)}^{\frac{3}{4}}}+c$
is the required solution.