JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Differential Equations

Let

f (x) = \int_{0}^{x} e^{t} f (t) d t + e^{x}

be a differentiable function for all

x \in R

. Then

f (x)

equals :

Answer: (A)

Solution

Differentiating

f (x) = \int_{0}^{x} e^{t} f (t) d t + e^{x}

gives

f^{'} (x) - e^{x} f (x) = e^{x}

and putting

x = 0

in the defining relation gives

f (0) = 1

. The integrating factor is

e^{- e^{x}}

, so

\frac{d}{d x} [f (x) e^{- e^{x}}] = e^{x} e^{- e^{x}} \Rightarrow f (x) e^{- e^{x}} = - e^{- e^{x}} + C

x = 0

this reads

e^{- 1} = - e^{- 1} + C

, so

C = 2 e^{- 1}

and

f (x) = e^{e^{x}} (2 e^{- 1} - e^{- e^{x}}) = 2 e^{e^{x} - 1} - 1

Thus

f (x) = 2 e^{(e^{x} - 1)} - 1

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