I know that $f(x)=e^x$ is the accepted and useful solution to $f'(x)=f(x)$, but why isn't $f(x)=0$ ever mentioned as a solution as well? Is it simply because it's not useful?
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2$f(x)=Ce^x$ for all constants $C$ are all the solutions. – Culver Kwan Jun 01 '20 at 01:36
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probably. Note that the ODE gives $a e^x$ where $a$ is a constant. This then includes $a=0.$ – Will Jagy Jun 01 '20 at 01:36
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1It's called the trivial solution for a reason. It's boring. – Sean Roberson Jun 01 '20 at 01:37
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@SeanRoberson, I had never even seen it as a trivial solution with a dismissal, which I always find surprising when I come across it. – Chuck Jun 01 '20 at 01:38
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1The source which mentions $e^x$ as solution of $f'=f$ does not mention $0$ as a solution of the differential equation for the same reason that it doesn't mention $2e^x$. – Jun 01 '20 at 01:38
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@Chuck That's the joke. The solution $f(x) = 0$ gives nothing interesting. – Sean Roberson Jun 01 '20 at 02:15
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Since $f'(x) = f(x)$ is linear differential equation, if $f(x)$ is solution then so is $kf(x)$ for $k \in \mathbb{R}$. So the solution is of the form $f(x) = Ae^x$ for $A \in \mathbb{R}$. Here $A$ is determined by some given initial condition.
The choice of $A =0$ gives the solution $f(x) = 0$.
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