Mathematics Differential Equations Linear first-order Differential Equation | Integrating-Factor

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# Integrating-Factor [mathjax]

Integrating-factor in ‘Linear first-order Differential Equation’ is a function, that makes the equation as a recognizable or exact derivative which is easy to solve simply by integration. To solve the separable equations, we multiplied both sides by an appropriate integrating factor $\frac{1}{y}$ which gave, $$\frac{1}{y}.\frac{dy}{dx}=2x \text{ that is, } D_{x}(ln(y)) = D_{x}(x^{2})\quad\text{ __Eqn(1)}$$

Here, each side of the eqn(1) is recognizable as a Derivative, all that remains are two simple integrations, which yield, $$ln(y) =x^{2} +C$$

In this case the function (ρ(y)=1/y) is called the integrating factor for the original equation, $$\frac{dy}{dx}=2xy, \quad\text{ y>0 }$$

It is commonly used to solve ordinary differential equations, but is also used within multivariable calculus when multiplying through by an integrating factor allows an inexact differential to be made into an exact equation.

## Linear first-order Differential Equation

The differential equation $$y^{′}+P(x)y=Q(x)$$

is considered as a model of linear first order differential equation. It involves the first derivative of $y$ with respect to $x$. The above equation is also linear because the derivative $y^{′}$ and the dependent variable $y$ appear in the equation are raised to the first power $(y^{′})^{1}$.

## Method of Solving Linear First-order Differential Equation

Four steps to solve linear first-order differential equations:

(1). Find the appropriate integrating factor of given DE.
(2). Multiply both sides of the equation by the integrating factor.
(3). Next, recognize the left side of resulting equation as derivative of product, $$\frac{d}{dx}ρ(x)y(x)=ρ(x)Q(x)$$
(4). Finally, integrate the equation, $$ρ(x)y(x)=\int{ρ(x)}{Q(x)}dx$$

## Let See in Action

With the aid of the appropriate integrating factor, there is a standard technique for solving the linear first-order equation. Suppose our first order differential equation is $$y^{′}+P(x)y=Q(x)\quad\text{_____eqn(2)}$$

On an interval on which the coefficient functions $P(x)$ and $Q(x)$ are continuous. For this equation our integrating factor $$ρ(x)=e^{\int{P(x)}dx}$$ Now we will multiply each side of the eqn(2) by the integrating factor, $$e^{\int{P(x)}dx}.y^{′}+P(x).e^{\int{P(x)}dx}=Q(x).e^{\int{P(x)}dx}\quad\text{___eqn(3)}$$

Because, $$\frac{d}{dx}(\int{p(x)dx)}=P(x)$$

So this is clear that the left hand side of eqn(3) is the derivative of the product $y(x).e^{\int{P(x)dx}}$, so equation (3) is equivalent to, $$\frac{d}{dx}(y(x).e^{\int{P(x)dx}}) = Q(x).e^{\int{P(x)dx}}$$

integration of the both side of the equation gives $$y(x).e^{\int{P(x)dx}} = \int{Q(x).e^{\int{P(x)dx}}}dx+C$$

Finally solving for (y), we obtain the the general solution of linear first-order equation for eqn(2): $$y(x) = e^{-\int{P(x)dx}}[\int{(Q(x).e^{\int{P(x)dx}})dx}+C]$$.

Example: Find the solution of the differential equation $\frac{dy}{dx}-2xy = x$

Solution: Comparing the given equation with the model we get, $P(x) = -2x) and (Q(x) =x$. Now to solve the equation we need to multiply both sides by the integrating factor $I = e^{\int{(P(x))}dx}$. For the equation $$I = e^{\int{(-2x)}dx} = e^{-x^{2}}$$

Now multiplying $I$ both sides of the equation gives,

$$e^{x^{2}}(\frac{dy}{dx}-2xy) = x.e^{-x^{2}}\\ ⇒ e^{-x^{2}}.y^{′}-2x.e^{-x^{2}}.y = x.e^{-x^{2}}$$

Now Integrating with respect to $x$,

$$\int{(e^{-x^{2}}.y^{′}-2x.e^{-x^{2}}.y)}dx = \int{x.e^{-x^{2}}}dx \quad\text{ __eqn(4)}$$

this is clear that the left side of the equation is the derivative of the product $e^{-x^{2}}.y$ which is our integrating factor, $I=e^{-x^{2}}$ time $y$, therefore eqn(4) can be re-written as

$${\int(\frac{d}{dx}(e^{-x^{2}}.y)})dx =\int{x.e^{-x^{2}}}dx \\ ⇒ e^{-x^{2}}.y = \int{x.e^{-x^{2}}}dx$$

Let, $u=-x^{2}, =>\frac{du}{dx} = -2x, =>du=-2xdx, =>xdx=-\frac{1}{2}du$, substituting to above equation gives,

$$e^{-x^{2}}.y = -\frac{1}{2}\int{e^{u}}.du\\ ⇒ e^{-x^{2}}.y = -\frac{1}{2}.e^{u}+C\\ ⇒ e^{-x^{2}}.y = -\frac{1}{2}.e^{-x^{2}}+C \quad\text{ substitute }u=-x^{2}$$

Technically this is the solution of given differential equation.

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