Given Differential equation is:

⇒

⇒

We know that 1–cos²x = sin²x

⇒ ……(1)

Let us assume z = x– 2y

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

⇒

⇒

⇒ ……(2)

Substitute (2) in (1) we get,

⇒

⇒

⇒

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

⇒

We know that

⇒ sec²zdz = dx

Integrating on both sides we get,

⇒ ∫sec²zdz = ∫dx

We know that:

(1) ∫sec²xdx = tanx + C

(2) ∫adx = ax + C

⇒ tanz = x + C

Since z = x – 2y we substitute this,

⇒ tan(x–2y) = x + C

∴ The solution for the given Differential Equation is tan(x–2y) = x + C.