Given and when x = 2, y = 1

This is a first order linear differential equation of the form

Here, and Q = 2y

The integrating factor (I.F) of this differential equation is,

We have

[∵ m log a = log a^m]

∴ I.F = y^–1 [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

We know

⇒ xy^–1 = 2y + c

⇒ xy^–1 × y = (2y + c)y

∴ x = (2y + c)y

However, when x = 2, we have y = 1.

⇒ 2 = (2 × 1 + c) × 1

⇒ 2 = 2 + c

∴ c = 2 – 2 = 0

By substituting the value of c in the equation for x, we get

x = (2y + 0)y

⇒ x = (2y)y

∴ x = 2y²

Thus, the solution of the given differential equation is x = 2y²