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