BITSAT 2013 Paper

We have

dy

dx

=

f′(x)

f(x)

y -

y2

f(x)

⇒

dy

dx

-

f′(x)

f(x)

y = -

y2

f(x)

Divide by y2 : y−2

dy

dx

- y−1

f′(x)

f(x)

= -

1

f(x)

Put y−1 = z ⇒ −y−2

dy

dx

=

dz

dx

-

dz

dx

-

f′(x)

f(x)

(z) = −

1

f(x)

⇒

dz

dx

dz

dx

+

f′(x)

f(x)

(z) =

1

f(x)

I.F. = e∫

f′(x)

f(x)

dx = elog‌f(x) = f (x)
∴ The solution is
z 9f(x)) = ∫

1

f(x)

(f(x)) dx + c
⇒ y−1 (f(x)) = x + c ⇒ f (x) = y (x + c)