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I'm currently reading the research paper, Two Singular Diffusion Problems, by William Feller (1950). However, I don't understand how Feller derived the solution $(3.5)$ given equation $(3.4)$ in his research paper. More specifically, I don't understand how Feller solved $$dt=\frac{d\omega}{f(t)-cs\omega} \implies \frac{d\omega}{dt}=f(t)-cs\omega,$$ where $$\text{This is equation (3.4)}\,\,\,\,\,\,\,\,\,\,\,\,\,\, e^{-bt}\frac{as-b}{s}=C_1 \implies s=\frac{be^{-bt}}{ae^{-bt}-C_1},$$ and $a, b, C_1$ are constants with $b\neq0$ to get the solution$$\text{This is equation (3.5)}\,\,\,\,\,\,\,\,\,\,\,\, \omega = \left|C_1 - ae^{-bt}\right|^{c/a}\left\{C_2 + \int_{0}^{t}{\frac{f(\tau)d\tau}{\left|C_1 - ae^{-b\tau}\right|^{c/a}}}\right\},$$ where $C_2$ is a constant as well.

Please note that I have already verified this is true by differentiating it (and using the Fundamental Theorem of Calculus) but I don't understand how Feller derived it originally.

Can someone please explain to me in details or give me some hints regarding this? Any help will be greatly appreciated.

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  • $\begingroup$ May be that this kind of question is more suitable for math.stackexchange.com than here. $\endgroup$
    – Quantopik
    Commented May 2, 2015 at 13:45
  • $\begingroup$ @Quantopic , I have already posted it there but I got no answer yet. So, I thought I should give this a try as well since the theorem in this paper also has an application in Financial Mathematics. Therefore, someone here may have already read this paper and may be able to help me with it. :) $\endgroup$
    – Vincent
    Commented May 3, 2015 at 6:43

1 Answer 1

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For simplicity, we assume the necessary positivity, and then we can ignore the absolute signs. Note that \begin{align*} \big(C_1 - a e^{-bt} \big) d\omega = \big(C_1 - a e^{-bt} \big) f(t) dt + cbe^{-bt} \omega dt. \end{align*} That is, \begin{align*} \big(C_1 - a e^{-bt} \big)\,d\omega - cbe^{-bt} \omega \, dt= \big(C_1 - a e^{-bt} \big) f(t)dt. \end{align*} Then, \begin{align*} d\Big(\big(C_1-ae^{-bt} \big)^{-\frac{c}{a}} \, \omega\Big) &= -\frac{c}{a}\,\omega \big(C_1-ae^{-bt} \big)^{-\frac{c}{a}-1}\big(abe^{-bt}\big) dt + \big(C_1-ae^{-bt} \big)^{-\frac{c}{a}} d\omega\\ &=\big(C_1-ae^{-bt} \big)^{-\frac{c}{a}-1}\Big[\big(C_1-ae^{-bt} \big) d\omega- cbe^{-bt} \omega \, dt \Big]\\ &=\big(C_1-ae^{-bt} \big)^{-\frac{c}{a}}f(t)dt. \end{align*} The remaining is now obvious.

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