Question

(y-yx) dx + (x+xy) dy=0 solve ​

Answer 1

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

Given that

$\rm :\longmapsto\:(y - xy)dx + (x + xy)dy = 0$

can be rewritten as

$\rm :\longmapsto\:y(1 - x)dx + x(1 + y)dy = 0$

$\rm :\longmapsto\:y(1 - x)dx = - \: x(1 + y)dy$

$\rm :\longmapsto\:y(x - 1)dx \: = \: x(1 + y)dy$

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

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

$\rm :\longmapsto\:\displaystyle\int\tt\bigg(1 - \dfrac{1}{x} \bigg)dx = \:\displaystyle\int\tt\bigg(1 + \dfrac{1}{y} \bigg)dy$

$\rm :\longmapsto\:x - log(x) = y + log(y) + c$

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

Hence,

Solution of

$\rm :\longmapsto\:(y - xy)dx + (x + xy)dy = 0$

is

$\rm :\longmapsto\:x - log(x) = y + log(y) + c$