Let [f(x) = _0^x e^tf(t)dt + e^x ] be a differentiable function for all x[]R.

Question

Let $$f(x) = \int\limits_0^x {{e^t}f(t)dt + {e^x}} $$ be a differentiable function for all x$$\in$$R. Then f(x) equals :

Accepted Answer

$$f(x) = \int\limits_0^x {{e^t}f(t)dt + {e^x}} $$ .... (1)

Differentiating both sides w.r.t. x

$f'(x) = {e^x}.f(x) + {e^x}$ (Using Newton L:eibnitz Theorem)

$$ \Rightarrow {{f'(x)} \over {f(x) + 1}} = {e^x}$$

Integrating w.r.t. x

$$\int {{{f'(x)} \over {f(x) + 1}}dx = \int {{e^x}dx} } $$

$$ \Rightarrow \ln (f(x) + 1) = {e^x} + c$$

Put x = 0

ln 2 = 1 + c ($$ \because $$ f(0) = 1, from equation (1))

$$ \therefore $$ $$\ln (f(x) + 1) = {e^x} + \ln 2 - 1$$

$$ \Rightarrow f(x) + 1 = 2.\,{e^{{e^x} - 1}}$$

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

Solutions · 1