CGPET 2015 Solved Paper

Given differential equation is
(y‌log‌x−1)y‌dx=xdy
⇒

dy

dx

=(y‌log‌x−1)

y

x

=

y2‌log‌x

x

−

y

x

⇒

dy

dx

+

y

x

=

y2‌log‌x

x

⇒

1

y2

dy

dx

+

y−1

x

=

log‌x

x

.... (i)
Put y−1=v⇒−y−2

dy

dx

=

dv

dx

⇒y−2

dy

dx

=−

dv

dx

From Eq. (i), we have
−

dv

dx

+

v

x

=

log‌x

x

⇒

dv

dx

−

v

x

=−

log‌x

x

...(ii)
This is linear differential equation.
Here, IF=e∫(−

1

x

)‌dx=e−log‌x=

1

x

So, solution is v.IF=∫IF.Q‌dx+C
v.

1

x

=∫

1

x

(−

log‌x

x

)‌dx+C
⇒v.

1

x

=−∫

log‌x

x2

‌dx+C
= −[log‌x(

−1

x

)+∫

1

x

.

1

x

‌dx]+C
⇒

1

x.y

=

log‌x

x

+

1

x

+C[∵v=

1

y

]
⇒ 1=y[log‌x+1+Cx]
⇒ 1=y[log‌x+log‌e+Cx][∵1=log‌e]
⇒1=y[log‌e.x+Cx]