The solution of differential equation (1 + x)y dx + (1 - y)x dy = 0 is

Question

The solution of differential equation (1 + x)y dx + (1 - y)x dy = 0 is (A)  loge(xy) + x - y = C (B)  loge ( (x /y))+ x + y = C (C)  loge ( (x /y))- x + y = C (D)  loge (xy)- x + y = C

Tardigrade Admin · Accepted Answer

Given, (1 + x)y dx + (1 - y)x dy = 0 ⇒ ((1-y)/y)dy+((1+x)dx/x) = 0 ⇒ ∫ ((1/y)-1)+((1/x)+1)dx=0 ⇒ log y - y + log x = C ⇒ log (xy) + x - y = C

Q. The solution of differential equation $(1 + x) y d x + (1 - y) x d y = 0$ is

$lo g_{e} (x y) + x - y = C$

$lo g_{e} (\frac{x}{y}) + x + y = C$

$lo g_{e} (\frac{x}{y}) - x + y = C$

$lo g_{e} (x y) - x + y = C$

Given, $(1 + x) y d x + (1 - y) x d y = 0$
$\Rightarrow \frac{( 1 - y )}{y} d y + \frac{( 1 + x ) d x}{x} = 0$
$\Rightarrow \int (\frac{1}{y} - 1) + (\frac{1}{x} + 1) d x = 0$
$\Rightarrow l o g y - y + l o g x = C$
$\Rightarrow l o g (x y) + x - y = C$

Q. The solution of differential equation (1+x)ydx+(1−y)xdy=0 is

loge​(xy)+x−y=C

loge​(yx​)+x+y=C

loge​(yx​)−x+y=C

loge​(xy)−x+y=C