Solve the differential equation given that when x = 2, y = 1.

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 am]

I.F = y–1 [ elog 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

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