Solve the differential equation `(dy)/(dx)=(x(2logx+1))/((siny+ycosy))`.

Explanation

`(dy)/(dx)=(x(2logx+1))/((siny+ycosy))`
`implies(siny+ycosy)dy=(2xlogx+x)dx`
`impliesintsinydy+intycosydy`
`=2intxlogxdx+intxdx+c`
`implies-cosy+[yintcosydy-int{(d)/(dy)y*intcosydy}dy]`
`=2[logxintxdx-int{(d)/(dx)logxintxdx}dx]+(x^(2))/(2)+c`
`implies-cosy+ysiny-int1*sinydy`
`=2[logx*(x^(2))/(2)-int(1)/(x)*(x^(2))/(2)dx]+(x^(2))/(2)+c`
`implies-cosy+ysiny+cosy`
`=x^(2)logx-intxdx+(x^(2))/(2)+c`
`impliesy siny=x^(2)logx-(x^(2))/(2)+(x^(2))/(2)+c`
`impliesysiny=x^(2)logx+c`.