x(logx) dy/dx + y =2(logx).......plz help me to how to solve it???????

Xlogx dy/dx+y 2/x logx Dividing by Xlogx --- Dy/dx + (1/Xlogx)y=2/x^2 Comparing with dy/dx+py=Q---- P=1/(Xlogx), Q=2/x^2 Now IF= e^[int(1/Xlogx)dx Let logx=u du=1/x dx xdu=dx So IF=e^[int{(1/xu).xdu}] IF=e^[int(1/u)du] IF=e^logu=solution of the equation Y×IF=[int(Q×IF)dx]+c ylogx=[int(2/x^2 × logx)dx]+c Let I = 2[int(logx × x^-2)dx]+c or, I=2[-logx/x+ int1/x^2dx] or, I=2[lox/x+ x^-1/(-1)] or, I = -2/x(1+logx) Putting value of I in the equation ylogx=-2/x(1+logx)+c

Hello hi madam