CGPET 2018 Solved Paper

Given differential equation,
x

dy

dx

=y(log‌y−log‌x+1)
⇒

dy

dx

=

y

x

(log‌y−log‌x+1)
⇒

dy

dx

=

y

x

(log‌

y

x

+1) ........(i)
It is homogeneous differential equation.
Now, put y=vx⇒

dy

dx

=v+x

dv

dx

........(ii)
From Eqs. (i) and (ii), we get
v+x

dv

dx

=v(log‌v+1)
⇒

dv

v‌log‌v

=

dx

x

By integrating both the sides, we have
log(log‌v)=log‌x+log‌C
⇒ log(log‌

y

x

)=log‌C‌x
⇒ log‌

y

x

=Cx
⇒ y=xeα