Question

Anyone slive this question​

Answer 1

$\large\underline{\sf{Solution-}}$

Given Differential equation is

$\rm :\longmapsto\:\dfrac{dy}{dx} + y = {e}^{ - x}$

Its a linear differential equation.

So,

On comparing with

$\rm :\longmapsto\:\dfrac{dy}{dx} + py = q \: where \: p \: and \: q \: \in \: f(x)$

So, we get

$\red{\rm :\longmapsto\:p = 1}$

$\red{\rm :\longmapsto\:q = {e}^{ - x} }$

Now,

Integrating Factor is

$\rm :\longmapsto\:I.F. \: = \: {e} \: \: ^{ \displaystyle \int \: p \: dx}$

$\rm :\longmapsto\:I.F. \: = \: {e} \: \: ^{ \displaystyle \int \: 1 \: dx}$

$\bf :\longmapsto\:I.F. \: = \: {e} ^{ x}$

Thus,

Solution is given by

$\rm :\longmapsto\:y \times I.F. = \displaystyle \int \: (q \times I.F.)dx$

$\rm :\longmapsto\:y \times {e}^{x} = \displaystyle \int \: {e}^{ - x} \times {e}^{x} \: dx$

$\rm :\longmapsto\:y \times {e}^{x} = \displaystyle \int \: 1 \: dx$

$\rm :\longmapsto\:y \times {e}^{x} = x + c$

$\bf\implies \:y {e}^{x} - x = c$

Basic Concept Used :-

Linear Differential equation :-

The form of the equation is

$\rm :\longmapsto\:\dfrac{dy}{dx} + py = q \: where \: p \: and \: q \: \in \: f(x)$

To solve this Differential equation, the following steps have to follow :

Step :- 1 Integrating Factor

$\red{\rm :\longmapsto\:I.F. \: = \: {e} \: \: ^{ \displaystyle \int \: p \: dx} }$

Step :- 2 Solution is given by

$\red{\rm :\longmapsto\:y \times I.F. = \displaystyle \int \: (q \times I.F.)dx}$

Answer 2

heya

can I know from where u got these solutions which is in attachment ??????