Let $I$ be an open interval and $C^{\infty}=C^{\infty}(I,\mathbb C)$. Does $f'-af=g$ have a solution for all $g\in C^{\infty}$ and $a\in\mathbb C$? Can we explicitely construct a solution?
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5Enormous hint - the LHS is $e^{at}(fe^{-at})'$ – Calvin Khor Nov 10 '20 at 07:43
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The good ole "method of integrating factors"! – Sangchul Lee Nov 10 '20 at 07:52
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Introduce $\tilde{f}(t):=f(t)e^{-at}$ and $\tilde{g}(t):=g(t)e^{-at}$. $f'-af=g$ and $\tilde{f}'=\tilde{g}$ are equivalent. $\tilde{f}(t)=\int_a^t\tilde{g}(x),\mathrm{d}x$ is a solution of $\tilde{f}'=\tilde{g}$, so $f(t)=\tilde{f}(t)e^{at}$ is a solution of $f'-af=g$. – Filippo Nov 10 '20 at 08:05
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Is that what you meant @CalvinKhor? – Filippo Nov 10 '20 at 08:06
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@Filippo yup, that's exactly what I meant (in fact you can add the constant of integration and give all solutions) – Calvin Khor Nov 10 '20 at 08:10
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@CalvinKhor Thank you for the confirmation and thank you for the comment (that we can construct the set of all solutions). – Filippo Nov 10 '20 at 08:17
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This speaks in favour of my alternative proof that homogeneous linear differential equation of order n have an n-dimensional solution space :) – Filippo Nov 10 '20 at 08:17