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