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Question

2 cos xdydx+4y sin x = sin 2x, given that y = 0 when x = π3.

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Solution

We have,
2cos xdydx+4y sin x=sin 2x
dydx+4ysin x2 cos x=2sin x cos x2 cos xdydx+2y tan x=sin x
Comparing with dydx+Py=Q, we getP=2tan xQ=sin xNow,I.F.=e2tan x dx=e2logsec x=sec2 xSo, the solution is given byy×I.F.=Q×I.F. dx + Cy sec2 x=sin x sec2 x dx + Cy sec2 x=tan x sec x dx + Cy sec2 x=sec x + Cy=cos x+C cos2x .....1Now, When x=π3, y=0 0=cos π3 + C cos2 π30=12+C14C=-2Putting the value of C in 1, we gety=cos x-2cos2 x

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