Question

(x+y-1)dx+(2x+2y+1)dy=0

Answer 1

We have,

-$(x + y -1)dx = (2x + 2y +1)dy$

$\dfrac{dy}{dx} = -\dfrac{(x+y-1)}{(2x+2y+1)}$

Let $x+y = t$

Differentiating with respect to 'x'

$1 + \dfrac{dy}{dx} = \dfrac{dt}{dx}$

$\dfrac{-(t-1)}{2t+1} = \dfrac{dt}{dx} - 1$

$1 - \dfrac{(t -1)}{(2t+1)} = \dfrac{dt}{dx}$

$\dfrac{dt}{dx} = \dfrac{(2t+1-t+1)}{(2t+1)}$

$\dfrac{(t+2)}{(2t+1)} = \dfrac{t+2}{2(t+\dfrac{1}{2})}$

$2\dfrac{dt}{dx} = \dfrac{t+2}{t+\dfrac{1}{2}}$

$\int\dfrac{(2t+1)}{t+2}dt = \int dx$

$\int \dfrac{(2t+4-3)}{(t+2)}dt = \int dx$

$\int 2dt -3 \int \dfrac{dt}{t+2} = \int dx$

$2t - 3ln|t+2| = x+c$

$2(x+y) - 3ln|x+y+2| = x+c$