The solution of the differential equation x3dydx+4x2tany=ex secy satisfying y1=0 is

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tany=(x-2)exlogx

siny=ex(x-1)x-4

tany=(x-1)exx-3

siny=ex(x-1)x-3

answer is B.

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Detailed Solution

x4cosydy+4x3sin⁡y=xex→∫dx4siny=∫xexdx⇒x4sin⁡y=ex(x−1)+c−⋯−−−1Pass⁡(1,0)⇒c=0→x4sin⁡y=ex(x−1)→sin⁡y=ex(x−1)x4

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