This is a linear differential equation of the form ‌
dy
dx
+P(x)y=Q(x), Where P(x)=‌
1
x‌log‌x
and Q(x)=‌
2
x
Now, IF=e∫‌
1
x‌log‌x
‌dx Let u=log‌x. Then du=‌
1
x
‌dx So, IF=e∫‌
1
u
du=elog‌u=u=log‌x So, the solution is given by ‌y⋅IF=∫Q(x)⋅IF‌dx+c ‌y‌log‌x=∫‌
2
x
‌log‌x‌dx+c Let v=log‌x Then, dv=‌
1
x
‌dx ‌‌ So, ‌y‌log‌x=∫2vdv+c=v2+c ‌⇒y‌log‌x=v2+c=(log‌x)2+c Given, y(e)=0 So, y‌log‌x=(log‌x)2+c ‌⇒0⋅log(e)=(log‌e)2+c ‌⇒‌‌0=1+c⇒c=−1 ∴ The particular solution is y‌log‌x=(log‌x)2−1 ⇒‌‌y=‌