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Question

dydx=sin x+x cos xy2 log y+1

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Solution

We have,
dydx=sin x+x cos xy2 log y+1y 2 log y+1dy= sin x+x cos xdxIntegrating both sides, we get2yII log y Idy+y dy=sin x dx+x cos x dx2log yy dy-2ddylog y×y dydy+y dy=-cos x+x cos x dx-dxdx×cos x dxy2log y-y dy+y dy=-cos x+x sin x dx+cos x+Cy2log y=x sin x+C

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