# Find the pa

cos y dx + (1+2e-x) sin y dy = 0

(1+2e-x) sin y dy = -cos y dx

On integrating, both sides

Taking 2 + ex = t in L.H.S,

ex dx = dt

Also, in R.H.S,

Taking sin y = u

cos y dy = du

ln t = ln u + ln c

As, t = 2 + ex, u = cos y

Therefore, ln (ex + 2) = - ln |cos y| + ln c

ln (ex + 2) = - ln |cos y| + ln c

We know that

As,

ln (ex + 2) = ln |sec y| + ln c

As, ln A + ln B = ln AB

So, ln (ex + 2) = ln |sec y| c

|ex + 2| = c|sec y|

ex + 2 = k sec y … (1)

Substituting, x=0

is the particular solution.

OR

Clearly, (1) is a Linear differential equation of the form,

With and Q (y) = 1

Solution of a Linear differential equation is given as,

I.F is known as integration factor and given by

= e(ln y + ln sin y)

I.F = eln(y sin y)

I.F = y sin y

Solution of the D.E. is:

xy sin y = -y cos y + sin y + c

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