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Let $f_h$ be the Hilbert transform of the real function $f$. I need some help solving this integro differential equation : $$\alpha f_h(x) + \beta f''(x) = f(x)$$

A simple sinusoid doesn't seem to fit the bill. Kind of puzzzling for me to even guess anything!

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  • $\begingroup$ $f(x) = Ce^{-kx}$ ?, $k$ is real. $\endgroup$ – Rajesh Dachiraju Dec 31 '13 at 10:28
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$$\alpha f_h(x) + \beta f''(x) = f(x)$$

Search for a particular solution on the form : $\quad f(x)=e^{i\omega x}$

$f_h(x)=i\,e^{i\omega x} \quad\to\quad \alpha\, i\,e^{i\omega x}-\beta\, \omega^2e^{i\omega x}=e^{i\omega x} \quad\to\quad i\,\alpha-\beta\,\omega^2=1$ $$i\,\omega=\left(\frac{1-i\,\alpha}{\beta} \right)^{1/2}$$ $$i\,\omega=\rho\left(\cos(\theta)+i\,\sin(\theta) \right)\qquad \begin{cases}\rho=\sqrt{\frac{\sqrt{1+\alpha^2}}{\beta}}\\ \theta=\frac{1}{2}\cot^{-1}(\alpha) \end{cases}$$ $$f(x)=e^{\rho\left(\cos(\theta)+i\,\sin(\theta) \right)\,x}$$

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