MHTCET 2016 Math Solved Paper

(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 =

1

e

Hence, x log x =

y

e

⇒ y = ex log x