Q. 214.5( 6 Votes )

Find the pa

Answer :

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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