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. |

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.

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

__Step 1:__**Divide the equation by $y^{n}$.**

__Step 2:__**Put $v = y^{1-n}$. Differentiating we get, $v'= (1-n)y^{n}y'$.**

__Step 3:__**Substituting the values of v and v' in the Bernoulli's equation we have**

__Step 4:__$\frac{1}{1-n}$ $v'+p(x)v=q(x)$

**Solve the differential equation obtained in step 4 to get the value of v.**

__Step 5:__**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**

__Step 6:__$y' +xy=xy^{2}$

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:

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

**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

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