Given Differential equation is:

⇒ ……(1)

Let us assume z = x + y + 1

Differentiating w.r.t x on both the sides we get,

⇒

⇒

⇒

⇒ ……(2)

Substituting (2) in (1) we get,

⇒

⇒

Bringing like variables on same (i.e, variable seperable technique) we get,

⇒

Integrating on both sides we get,

⇒

We know that and

Also ∫adx = ax + C

⇒

⇒ tan^–1z = x + C

We know that z = x + y + 1 , substituting this we get,

⇒ tan^–1(x + y + 1) = x + C

∴ The solution for the given Differential equation is tan^–1(x + y + 1) = x + C