What class of functions $f(x)$ satisfies $f'(x)+f(x)=k$?
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is your k constant? – Yimin May 11 '15 at 23:53
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1how about multiply $e^x$ on both side, and try to diff $e^x f(x)$. – Yimin May 11 '15 at 23:53
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1An elementary solution to a broader class of differential equations can be found here. – Git Gud May 11 '15 at 23:56
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$(e^x f)' = e^x (f + f') = e^x k$.
$e^x f = \int^x_0 e^s k(s) + C$
$f = e^{-x} \int_0^x e^s k(s) ds + Ce^{-x}$
Yimin
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I suppose that $ k$ is a real number. If yes, then you multiply both sides by $e^x$ and use that:
If $f'(x) =g'(x), \quad \forall x \in I$, where $I \subset \mathbb R $ is an interval, then
$$ f(x) = g(x) + c , \quad x \in I$$
where $c \in \mathbb R $ is a constant.
Namely, in your case we have:
$ f'(x)+ f(x)= k \Longleftrightarrow e^x f'(x) + e^x f(x) = ke^x \Longleftrightarrow \left(e^x f(x) \right)' = (ke^x)' $
Thus, $e^x f(x) = k e^x + c, \quad x \in \mathbb R$, and $c$ is a real constant. Hence, you obtain:
$$ f(x) = ce^{-x} + k \quad x \in \mathbb R $$
passenger
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