VITEEE 2025 Paper

The given differential equation is
‌(1+e‌

x

y

)‌dx+e‌

x

y

(1−‌

x

y

)dy=0
‌‌

dx

dy

=‌

e x y ( x y −1)

(ex∕y+1)

‌=g(‌

x

y

)∵‌

dx

dy

=g(‌

x

y

)
∴ eq. (i) is the homogeneous differential equation so, put
‌

x

y

=v‌ i.e., ‌x=vy⇒‌

dx

dy

=v+y‌

dv

dy

Then, eq. (i) becomes
‌v+y‌

dv

dy

=‌

ev(v−1)

ev+1

⇒y‌

dv

dy

=‌

ev(v−1)

ev+1

−v
‌⇒y‌

dv

dy

=‌

vv−ev−vev−v

ev+1

−v
‌⇒‌

ev+1

ev+v

dv=−‌

1

y

dy
On integrating both sides, we get
‌∫‌

ev+1

ev+v

dv=−∫‌

1

y

dy
⇒ev+1=‌

dt

dv

⇒dv=‌

dt

ev+1

∴‌∫‌

ev+1

t

‌

dt

ev+1

−log|y|+log‌C
⇒log|t|+log|y|=log‌C
⇒log‌|ev+v|+log|y|=log‌C
⇒log‌|(ev+v)y|=C⇒|(ev+v)y|=C
⇒(ev+v)y=C.‌ So, put ‌v=‌

x

y

,‌ we get ‌
‌(ex∕y+‌

x

y

)t=ev+v)
‌
This is the required solution of the given differential equation.