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Question

Find the particular solution of the differential equation ex tan y dx+(2ex)sec2y dy=0, given that y=π4 when x=0.

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Solution

The given equation is
extan y dx+(2ex)sec2y dy=0

ex2exdx=sec2ytan ydy

ex2exdx=sec2ytan ydy

Put 2ex=zexdx=dz

And tany=u sec2y dy=du
1zdz=1udu

log z=log u+log C [C is arbitary constant of intregation]

log z=log (uC)

z=uC

2ex=Ctan y ....... (i)

When x=0, then y=π4 [given]

2e0=Ctanπ4

21=C1C=1

putting C=1 in the equation (i)

The particular solution of the given differential equation is,

tan y=2ex

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