Solution of (x + y - 1)dx + (2x + 2y - 3)dy = 0 is: |
y + x + log (x + y - 2) = c y + 2x + log (x + y - 2) = c 2y + x + log (x + y - 2) = c 2y + 2x + log (x + y - 2) = c |
2y + x + log (x + y - 2) = c |
$\frac{dy}{dx}=-\frac{x+y-1}{2x+2y-3}$ Put x + y = t; $1+\frac{dy}{dx}=\frac{dt}{dx};\frac{dt}{dx}-1=-\frac{t-1}{2t-3}$ $\frac{dt}{dx}=\frac{t-2}{2t-3};\int\frac{(2t-3)dt}{t-2}=\int dx$; 2t + loge (t - 2) = x + c; x + 2y + loge (x + y - 2) = c |