UPSEE 2017 Solved Paper

dy

dx

=

y2−x

2y(x+1)

........(i)
⇒y

dy

dx

=

y2−x

2(x+1)

⇒y

dy

dx

−

y2

2(x+1)

=−

x

2(x+1)

.......(ii)
Put y2=t
By differentiating both side w.r.t. 'x', we get
2y

dy

dx

=

dt

dx

∴ Eq. (ii) reduces to

1

2

dt

dx

−

t

2(x+1)

=

−x

2(x+1)

⇒

dt

dx

−

1

x+1

t=

−x

x+1

This is linear differential equation with
P=

−1

x+1

and Q=

−x

x+1

∴IF=e∫Pdx
=e∫

−1

(x+1)

dx
=e−log(x+1)=elog‌

1

x+1

=

1

x+1

∴ Required solution will be
t. IF=∫Q(IF)dx+logC[ Here, log‌c is constant ]
y2.

1

1+x

=∫(

−x

x+1

×

1

x+1

)dx +log‌c
⇒

y2

1+x

=−∫

(x+1)−1

(x+1)2

dx +log‌c
⇒

y2

1+x

=−[∫

1

1+x

dx−∫

1

(1+x)2

dx] +log‌c
⇒

y2

1+x

=−[log(1+x)+

1

1+x

] +log‌c
⇒

y2

1+x

=log‌

c

1+x

−

1

1+x

⇒y2=(1+x)‌log‌

c

1+x

−1