Differential Equations

It is given that
6

x

∫

1

f(t)dt = 3xf(x)−x3
⇒ 6 f (x) = 3 f (x) + 3 xf' (x) - 3x2
⇒ 3 f (x) = 3 xf' (x) - 3x2 ⇒ xf' (x) - f (x) = x2
⇒ x

dy

dx

−y = x2 ⇒

dy

dx

−

1

x

y = x (1)
Now, I.F. = e∫−

1

x

dx = e−logex
Multiplying Eq. (1) both sides by

1

x

, we get

1

x

dy

dx

−

1

x2

y = 1 ⇒

d

dx

(y.

1

x

) = 1
Integrating, we get

y

x

= x + c
Substituting x = 1 and y = 2, we get
⇒ 2 = 1 + c ⇒ c = 1 ⇒ y = x2+x
⇒ f (x) = x2+x ⇒ f (2) = 6
Note: If we put x = 1 in the given equation, we get f (1) = 1/3.
Question is ambiguous as 8/3 can also be the answer.