Given Differential equation is:

⇒ ……(1)

Let us assume z = x + y

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

⇒

⇒

⇒ ……(2)

Substituting (2) in (1) we get,

⇒

⇒

Bringing the like variables to same side (i.e., Variable seperable technique) we get,

⇒

Integrating on both sides we get,

⇒

⇒

We know that:

(1)

(2)

⇒

⇒ tan^–1z = x + C

Since z = x + y we substitute this,

⇒ tan^–1(x + y) = x + C

⇒ x + y = tan(x + C)

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