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Question

Solve the differential equation by variable separable method:(ey+1)cosxdx+eysinxdy=0.

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Solution

Consider given the deferential equation,


(ey+1)cosxdx+eysinxdy=0

(ey+1)cosxdx=eysinxdy

cosxdxsinx=eyey+1dy

cotx.dx=eyey+1dy


Taking integration both sides,

cotx.dx=eyey+1dy

logsinx=log(ey+1)+logC

logsinx=logC(ey+1)

sinx=C(ey+1)

(ey+1)sinx=C


Hence, this is the answer.

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