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# Find the particular solution of the differential equation y ≠ 0 given that x = 0 when **[CBSE 2013]**

**[CBSE 2013]**

Answer :

Given

This is a first order linear differential equation of the form

Here, P = cot y and Q = 2y + y^{2}cot y

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

We have

∴ I.F = sin y [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Recall

⇒ x sin y = y^{2} sin y + c

∴ x = y^{2} + c cosec y

However, when, we have x = 0.

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

Thus, the solution of the given differential equation is

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