dy/dx = 1/2 y^2 cot x (ii) Given that y = 2 when x = 1/6 π, solve this differential equation to find the equation of the curve.?

dy/dx = (1/2)y²cotx

so, ∫ (1/y²) dy = ∫ cotx dx

=> (-1/y) = ln(sinx) + C

Now, with y = 2 and x = π/6 we have:

-1/2 = ln[sin(π/6)] + C

i.e. -1/2 = ln(1/2) + C

so, C = -1/2 - ln(1/2)

Hence, (-1/y) = ln(sinx) -1/2 - ln(1/2)

or, 1/y = 1/2 + ln(1/2) - ln(sinx)

=> 1/y = 1/2 + ln[1/(2sinx)]

=> 2/y = 1 + 2ln[1/(2sinx)]

i.e. y/2 = 1/{1 + 2ln[1/(2sinx)] }

Hence, y = 2/{1 + 2ln[1/(2sinx)]}.....yuk!!

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