Question

the general solution of dy / dx +
xy = x is​

Answer 1

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

$\red{\rm :\longmapsto\:\dfrac{dy}{dx} + xy = x}$

can be rewritten as

$\rm :\longmapsto\:\dfrac{dy}{dx} = x - xy$

$\rm :\longmapsto\:\dfrac{dy}{dx} = x(1 - y)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = - x(y - 1)$

$\rm :\longmapsto\:\dfrac{dy}{y - 1} = - x \: dx$

$\displaystyle\rm :\longmapsto\: \int\dfrac{dy}{y - 1} = - \int \: x \: dx$

$\rm :\longmapsto\: log(y - 1) = - \dfrac{ {x}^{2} }{2} + c$

$\red{\bigg \{ \because \: \displaystyle \int \: \dfrac{1}{x}dx = log(x) + c \: \bigg \}} \\ \red{\bigg \{ \because \displaystyle \int \: {x}^{n}dx = \dfrac{ {x}^{n + 1} }{n + 1} + c\: \bigg \}}$

$\rm :\longmapsto\: log(y - 1)^{2} = - {x}^{2} +2 c$

$\red{\bigg \{ \because \: log( {x}^{y} ) = y \: logx \: \bigg \}}$

$\rm :\longmapsto\: log(y - 1)^{2} + {x}^{2} = d \: \: \: where \: d = 2c$

Additional Information :-

Solution of linear differential equation :-

The linear differential equation is of the form

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

Step :- Integrating Factor

$\rm :\longmapsto\:Integrating \: factor, \: i.f. \: = {e} \: ^{ \displaystyle \int \: pdx}$

Step :- 2 Solution is given by

$\rm :\longmapsto\:y \times i.f. \: = \: \displaystyle \int \: (q \times i.f.) \: dx$