If f(x) = f'(x) + f''(x) + f'''(x) + ldots an | vedclass

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(A) Given the equation: $f(x) = f'(x) + f''(x) + f'''(x) + \ldots$ Assume $f(x) = e^{kx}$. Then $f'(x) = ke^{kx}$,$f''(x) = k^2e^{kx}$,$f'''(x) = k^3e

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$e^{x / 2}$

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(A) Given the equation: $f(x) = f'(x) + f''(x) + f'''(x) + \ldots$ Assume $f(x) = e^{kx}$. Then $f'(x) = ke^{kx}$,$f''(x) = k^2e^{kx}$,$f'''(x) = k^3e^{kx}$,and so on. Substituting these into the given equation: $e^{kx} = ke^{kx} + k^2e^{kx} + k^3e^{kx} + \ldots$ Dividing by $e^{kx}$ (since $e^{kx} 
eq 0$): $1 = k + k^2 + k^3 + \ldots$ This is an infinite geometric series with first term $a = k$ and common ratio $r = k$. The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$ for $|r| So,$1 = \frac{k}{1-k}$. $1 - k = k \implies 2k = 1 \implies k = \frac{1}{2}$. Thus,$f(x) = Ce^{x/2}$. Given $f(0) = 1$,we have $Ce^0 = 1$,which implies $C = 1$. Therefore,$f(x) = e^{x/2}$.

If $f(x) = f'(x) + f''(x) + f'''(x) + \ldots$ and $f(0) = 1$,then $f(x)$ is equal to

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