Question

Solve the question of B.SC.

dy/dx+ √(1-y²/1-x²)=0​

Answer 1

Answer:

ANSWER

Given differential eqn can be written as

dx

dy

=−

xy

(x

2

−1)(y

2

−1)

⇒

y

2

−1

y

dy=−

x

x

2

−1

dx

Integrating both sides

∫

y

2

−1

y

dy=−∫

x

x

2

−1

dx

Put y

2

−1=t

⇒2ydy=dt

Also, put

x

2

−1

=u

⇒x

2

−1=u

2

⇒xdx=udu

∫

2

t

1

dt=−∫

1+u

2

u

udu

⇒∫

2

t

1

dt=−∫

1+u

2

u

2

du

⇒

t

=−∫(1−

1+u

2

1

)du

⇒

y

2

−1

=−u+tan

−1

u+C

⇒

y

2

−1

=−

x

2

−1

+tan

−1

x

2

−1

+C

⇒

y

2

−1

=−

x

2

−1

+sec

−1

x+C (tan

−1

x

2

−1

=z⇒

x

2

−1

=tanz⇒secz=x )