(a) The given differential equation is y(1+log‌x)[
dx
dy
] - x‌log‌x=0 ⇒
(1+log‌x)dx
x‌log‌x
=
dy
y
⇒ (
1
x‌log‌x
+
1
x
)dx=
1
y
dy After integrating on both sides, we get ∫(
1
x‌logx
+
1
x
)dx = ∫
1
y
dy Let log‌x=t ⇒
1
x
dx=dt So, ∫
1
t
dt+∫
1
x
dx = ∫
1
y
dy ⇒ log t + log x = log y + log c ⇒ log tx = log yc ⇒ tx = yc ⇒ x log x = yc When x = e then y = e2 Therefore, e log e = e2c ⇒ e x 1 = e2c ⇒ c =