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Question

The solution of the differential equation dydx+y2 secx=tanx2y, where 0xπ2, and y(0)=1, is given by:

A
y2=1+xsecx+tanx
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B
y=1xsecx+tanx
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C
y=1+xsecx+tanx
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D
y2=1xsecx+tanx
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Solution

The correct option is D y2=1xsecx+tanx
dydx+y2 secx=tanx2y
2ydydx+y2 secx=tanx
Put y2=v 2ydydx=dvdx
dvdx+vsecx=tanx
I.F.=esecx dx
=eln(secx+tanx)=secx+tanx
The general solution is v (secx+tanx)=tanx (secx+tanx) dx
y2 (secx+tanx)=secx+tanxx+C
Given, y(0)=1C=0
y2=1xsecx+tanx

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