AP EAMCET Engineering 2018 Apr 22 Shift 1 Paper

It is given,
(1+ex∕y)dx+ex∕y(1−

x

y

)dy=0
(1+ex∕y)dx=−ex∕y(1−

x

y

)dy

dy

dx

=

−ex∕y(1− x y )

(1+ex∕y)

......(i)
Substituting x=vy and differentiating it with respect to y.

dx

dy

=v

dy

dy

+y

dv

dy

dx

dy

=v+y

dv

dy

Substitute these values in equation (i).
v+v

dv

dy

=

−ev(1−v)

(1+ev)

y

dv

dy

=

−ev(1−v)

(1+ev)

−v
y

dv

dy

=

−(v+ev)

(1+ev)

−

dy

y

=[

1+ev

v+ev

]dv
Integrating both sides we get,
−∫

dy

y

=∫[

1+ex

v+ev

]dv
∫

1+ex

v+ev

dv=−log‌y+log‌c
Put, v+ev=t
(1+ev)dv=dt
∴∫

dt

t

=−log‌y+log‌c
log‌t=−log‌y+log‌c
log‌y(v+ev)=log‌c
Put v as

x

y

in above.
log‌y(

x

y

+ex∕y)=log‌c
y (

x

y

+ex∕y)=c
x+ye

x

y

=c