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Question

The general solution of the differential equation, sin2x(dydxtanx)y=0, is

A
ytanx=x+c
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B
ycotx=tanx+c
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C
ytanx=cotx+c
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D
ycotx=x+c
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Solution

The correct option is D ycotx=x+c
sin2x(ytanx)+y=0

y=tanxysin2x

y+ysin2x=tanx

If =ef(x)dx=edxsin2x=

=esec2x 2tanxdx=edt2t where t=tanx......(solve the above integration we get)

=cotx

y(I.F.)=(I.F.)g(x)dx+c

ycotx=x+c

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