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Question

Let y=y(x) be the solution of the differential equation sinxdydx+ycosx=4x,xϵ(0,π). If y(π2)=0, then y(π6) is equal to

A
89π2
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B
49π2
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C
493π2
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D
893π2
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Solution

The correct option is A −89π2Given differential equation:sinxdydx+ycosx=4x where x∈(0,π)∴dydx+ycosxsinx=4xsinxdydx+cotxy=4xsinxComparing with dydx+P(x)y=Q(x)∴P(x)=cot(x),Q(x)=4xsinxI.F =e∫P(x)dx=e∫cotxdx=e∫cotxdx=e∫log(|sinx|)dx=sinx∴ solution is given byyI.F=∫Q(x)IFdx+cysinx=∫4xsinxsinxdx+c=∫4xdx+c =4x22+c =2x2+c∴ysinx=2x2+c..................(A)given that y(π2)=0i.e when x=π2,y=0Equation becomes : 0=2(π2)2+c0=2×π24+c∴c=π22put this value in (A) ∴ysinx=2x2−π22Next to find y(π6) take x=π6∴ysinπ6=2(π6)2−π22y=12=2.π236−π22y2=π218−π2×92×9y2=π2−9π218y=−8π29∴y(π6)=−89π2

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