The solution of the equation dydx=x(2log⁡x+1) siny+ycosy is

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ysiny=x2logx+x22+c

ycosy=x2logx+1+c

ycosy=x2logx+x22+c

ysiny=x2logx+c

answer is D.

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Detailed Solution

∫ycosy+sinydy=∫2x lnx+xdx∫ycosy dy+∫siny dy=2∫xlnx dx+∫x dx integrate only first termsy∫cosy dy-∫1 ∫cosy dydy+∫siny dy=2 lnx∫x dx-∫1x∫x dxdx+∫x dx ysiny-∫siny dy+∫siny dy=2lnx x22-∫1xx22dx+∫x dx y siny=x2ln x-22∫x dx+∫x dx ⇒ysiny=x2ln x+c

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