Question

3. (x + y − 1)dx + (2x + 2y + 1)dy = 0

Answer 1

We have given the differential equation,

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

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

Substituting $x+y = t$

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

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

Putting value in equation(1),

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

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

Separating terms and integrating both sides,

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

$t- log(t+1) + \dfrac{1}{2}log(1+t) = x + C$

$x+y-log(x+y+1)+ \dfrac{1}{2}log(1+t) = x+C$