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.