Find the particular solution of the following differential equation.

Cos y dx+(1+2e^-x) sin y dy=0; y(0)=π/4

OR

Find the general solution of the differential equation:

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 + e^x = t in L.H.S,

⇒ e^x dx = dt

Also, in R.H.S,

Taking sin y = u

⇒ cos y dy = du

⇒ ln t = ln u + ln c

As, t = 2 + e^x, u = cos y

Therefore, ln (e^x + 2) = - ln |cos y| + ln c

⇒ ln (e^x + 2) = - ln |cos y| + ln c

We know that

As,

⇒ ln (e^x + 2) = ln |sec y| + ln c

As, ln A + ln B = ln AB

So, ln (e^x + 2) = ln |sec y| c

⇒ |e^x + 2| = c|sec y|

⇒ e^x + 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 = e^{ln(y sin y)}

⇒ I.F = y sin y

∴ Solution of the D.E. is:

⇒ xy sin y = -y cos y + sin y + c