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# Bernoulli's Differential Equation

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 Sub Topics Differential equations are important part of calculus. A differential equation consists of a function and its derivatives. It can be classified on the basis of its linearity. The linear differential equation will have all the function and its derivatives with power of one. The non-linear differential equations have derivatives and the function having power of more than one. It is tough to solve a non-linear differential equation. To solve a non-linear differential equation, mostly it is being reduced to a linear differential equation. Bernoulli's equation is one such non-linear differential equation which is needed to be reduced to linear form for getting its solution.

## Equation

The differential equation $y' + p(x)y = Q(x)y^{n}$ is known as the bernoulli's equation. When n = 0, it becomes a linear differential equation, y' + p(x)y = Q(x). When n = 1, then also it becomes a linear equation y' + p(x)y = Q(x)y.

It is a non-linear equation for n > 1. To make such equation linear, the complete equation is divided by $y^{n}$ and then the substitution $v = y^{1-n}$ is used which results in a linear differential equation in terms of v. ]

To solve the linear differential equation of the form  y' + P(x)y = Q(x), we first find the integrating factor which is  $e^{\int P(x)dx}$.

Multiply the whole equation by the IF. Then, it will be of the form ((IF).y)'=Q(x).(IF). Integrating both the sides the solution comes in form of  $y(I.F)=\int Q(x)(I.F)$. Hence, the solution for y is obtained.

## Steps

Step 1: Bernoulli's equation is of the form $y' + p(x)y = Q(x)y^{n}$.

Step 2: Divide the equation by $y^{n}$.

Step 3: Put $v = y^{1-n}$. Differentiating we get, $v'= (1-n)y^{n}y'$.

Step 4:  Substituting the values of v and v' in the Bernoulli's equation we have

$\frac{1}{1-n}$ $v'+p(x)v=q(x)$

Step 5: Solve the differential equation obtained in step 4 to get the value of v.

Step 6:  Plug the substitution back to get the value in terms of y from the equation $v = y^{1-n}$.
Let us take an example and understand. Let us take a Bernoulli's differential equation

$y' +xy=xy^{2}$

We will divide the equation by $y^{2}$. Thus, we get the equation as

$y^{-2}y'+xy^{-1}=x$ ... (1)

For substitution we take, $v = y^{1-2}=y^{-1}$. Differentiating both sides, we have
$v'=(-1)y^{-2}y'$. Now, substitute these values of v and v' in equation (1).

v' - vx = -x

This becomes the general form of a linear differential equation in terms of v. We take the integrating factor and solve this equation. We get the solution as,

$v$=$\frac{1}{2}$+$ce^{x^{2}}$.

Substituting this value of v back in $v =y^{-1}$, we have solution as:

$\frac{1}{y^{2}}$=$\frac{1}{2}$ +$ce^{x^{2}}$

## Examples

Example 1: Solve the given Bernoulli's equation.

$y'-y=e^{-x}y^{-2}$

Solution: To make the equation linear, first divide it by $y^{-2}$. The resulting equation will be,

$y^{2}y'-y^{3}=e^{-x}$

Substitute using $v=y^{3}, v'=3y^{2}y'$. The equation will come in terms of v as,

$\frac{1}{3}$ $v'-v=e^{-x}\Rightarrow v'-3v=3e^{-x}$ ... (1)

As it becomes a linear differential equation get the integrating factor.

I.F = $e^{\int -3dx}=e^{-3x}$

Multiplying equation (1) by I.F,

$v'e^{-3x}-3ve^{-3x}=3e^{-x}e^{-3x}$

$\Rightarrow (ve^{-3x})'=3e^{-4x}$

Integrating both sides,

$(ve^{-3x})'$=$\frac{-3}{4}$ $e^{-4x}+c$

$\Rightarrow v=ce^{3x}$ -$\frac{3}{4}$ $e^{-x}$

As it has been taken $v=y^{3}$, putting this value the solution becomes

$y^{3}$ =$ce^{3x}$-$\frac{3}{4}$ $e^{-x}$

Example 2: Solve the equation $y'+y=y^{2}$.

Solution: The given equation is in the form $y' + p(x)y = Q(x)y^{n}$. Hence, this is a Bernoulli's equation.

Divide the whole equation by $y^{2}$. The resulting equation is,

$y^{-2}y'+y^{-1}=1$.

Substitute using $v= y^{-1}, v'=-y^{-2}y'$ and get the linear differential equation in terms of v.

$-v'+v=1\Rightarrow v'-v=-1$. ...(1)

Integrating factor = $e^{\int -dx}=e^{-x}$

Multiplying equation (1) by I.F,

$v'e^{-x}-ve^{-x}=-e^{-x}$

$\Rightarrow (ve^{-x})'=-e^{-x}$

Integrating both the sides,

$(ve^{-x})=e^{-x}+c$

v =  $ce^{x}+1$

Plugging back the substitution $v= y^{-1}$ the solution becomes

$\frac{1}{y}$ =$ce^{x}+1$