dy/dx=x(2logx+1)/siny+ycosy
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Sahdev Sharma 8 years, 4 months ago
{tex}{dy\over dx}={x(2log\ x+1)\over sin\ y+y.cos\ y}{/tex}
{tex} (sin \ y+y.cos\ y)dy = [x(2log\ x+1)]dx{/tex}
Integrating both sides, we get
{tex} \int sin\ y\ dy +\int y.cos\ y\ dy = \int 2x.log\ x \ dx+ \int xdx{/tex}
{tex}-cox \ y + y\ sin\ y + cos \ y = x^2log\ x -{x^2\over 2}+ {x^2\over 2}+C{/tex}
{tex}=> y sin\ y =x^2.log \ x+ C{/tex}
7Thank You