Solve the following differential equation: `(x+y-1)dy=(x+y)dx`

Explanation

`(x+y-1) d y=(x+y) d x `

`\frac{d y}{x+y}=\frac{d x}{x+y-1}=k \text { say }`

Then `k=\frac{d y+d x}{x+y+x+y-1}=\frac{d(x+y)}{2(x+y)-1}=\frac{1}{2} \frac{d(2 x+2 y-1)}{2 x+2 y-1}`

Also `k=\frac{d y-d x}{x+y-x-y+1}=\frac{d(y-x)}{1}`

So `d(y-x)=\frac{1}{2} \frac{d(2 x+2 y-1)}{2 x+2 y-1}`

Integrating,

`y-x=\frac{1}{2} \ln |2 x+2 y-1|+C`