Verified answer

[tex]\large\underline{\sf{Solution-}}[/tex]

Given differential equation is

[tex]\sf \: \dfrac{dy}{dx} = \dfrac{x - y + 1}{x + y + 1} \\ \\ [/tex]

can be rewritten as

[tex]\sf \: (x - y + 1)dx = (x + y + 1)dy \\ \\ [/tex]

[tex]\sf \: (x - y + 1)dx - (x + y + 1)dy = 0\\ \\ [/tex]

[tex]\sf \: (x - y + 1)dx + ( - x - y - 1)dy = 0\\ \\ [/tex]

[tex] \bf \: On \: comparing \: with \: Mdx + Ndy = 0, \: we \: get \\ \\ [/tex]

[tex]\sf \: M = x - y + 1 \\ \\ [/tex]

[tex]\sf \: N = - x - y - 1 \\ \\ [/tex]

Now, Consider

[tex]\bf \: \dfrac{\partial M}{\partial y} = \dfrac{\partial }{\partial y}(x - y + 1) = - 1 \\ \\ [/tex]

and

[tex]\bf \: \dfrac{\partial N}{\partial x} = \dfrac{\partial }{\partial x}( - x - y - 1) = - 1 \\ \\ [/tex]

[tex]\bf\implies \:\dfrac{\partial N}{\partial x} = \dfrac{\partial M}{\partial y} \\ \\ [/tex]

[tex]\bf\implies \:(x - y + 1)dx + ( - x - y - 1)dy = 0 \: is \: exact. \\ \\ [/tex]

So, Solution of differential equation is given by

[tex]\sf \: \displaystyle\int_{\sf \: y \: constant}\sf M \: dx \: + \displaystyle\int_{\sf \: term \: not \: containing \: x}\sf Ndy = c \\ \\ [/tex]

[tex]\sf \: \displaystyle\int_{\sf \: y \: constant}\sf (x - y + 1) \: dx \: + \displaystyle\int_{\sf \: term \: not \: containing \: x}\sf ( - y - 1)dy = c \\ \\ [/tex]

[tex] \sf \: \dfrac{ {x}^{2} }{2} - xy + x - \dfrac{ {y}^{2} }{2} - y = c \\ \\ [/tex]

[tex] \sf \: {x}^{2} - 2xy + 2x - {y}^{2} - 2y = 2c \\ \\ [/tex]

[tex]\implies \:\bf \: {x}^{2} - 2xy + 2x - {y}^{2} - 2y = c' \: \: \{where \: c' = 2c \}\\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]