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Question

Find the particular solution of the differential equation dydx+2y tan x=sin x, given that y=0 when x=π3.

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Solution

The given differential equation (D.E) is dydx+2y tan x=sin x

Comparing it with dydx+Py=Q, we have

P=2 tanx and Q=sinx

Now the integral factor (I.F) of this type of D.E is given by
I.F=eP.dx=e2 tan x dx=e2 log |sec x|=elog sec2x=sec2x

Solution of given differential equation is :

y (I.F)=Q (I.F) dx+C

y.sec2x=sin x.sec2x dx+C

y.sec2x=tan x.secx dx+C

y.sec2x=sec x+C

When x=π3, then y=0

0.sec2π3=secπ3+C

C=2

Hence, the required particular solution of the differential equation is

y.sec2x=secx2

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