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Question

Solve the differential equation dydx  3y cotx = sin2x, y = 2 when  x =π2 .


Solution

dydx + (-3 cot x)y = sin 2x

It is an L.D.E in y

P = -3 cot x, Q = sin 2x

I.F = epdx = e3logsinx=1sin3x

G.S: y (IF)  = Q (IF) dx + c

ysin3x = 2sinx cosxsin3x dx + c

y cosec3x = 2 (1sinx)+c .....(1)

(1) passes through (π2,2)

2 = - 2 + c c = 4

y cosec3 x = 2sinx + 4

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