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C̶o̶n̶s̶i̶d̶e̶r̶ ̶a̶ ̶R̶i̶c̶a̶t̶t̶i̶ ̶e̶q̶u̶a̶t̶i̶o̶n̶ ̶o̶f̶ ̶t̶h̶e̶ ̶f̶o̶r̶m̶

$$ y' + y^2 = S(x), \qquad \qquad \qquad (1)$$

w̶h̶e̶r̶e̶ ̶$̶S̶(̶x̶)̶$̶ ̶i̶s̶ ̶a̶ ̶m̶e̶r̶o̶m̶o̶r̶p̶h̶i̶c̶ ̶f̶u̶n̶c̶t̶i̶o̶n̶,̶ ̶a̶n̶d̶ ̶$̶y̶$̶ ̶i̶s̶ ̶a̶ ̶c̶o̶m̶p̶l̶e̶x̶-̶v̶a̶l̶u̶e̶d̶ ̶f̶u̶n̶c̶t̶i̶o̶n̶.̶ ̶D̶o̶e̶s̶ ̶t̶h̶e̶r̶e̶ ̶e̶x̶i̶s̶t̶ ̶a̶n̶y̶ ̶t̶r̶a̶n̶s̶f̶o̶r̶m̶a̶t̶i̶o̶n̶ ̶s̶u̶c̶h̶ ̶t̶h̶a̶t̶ ̶$̶(̶1̶)̶$̶ ̶c̶a̶n̶ ̶b̶e̶ ̶r̶e̶-̶w̶r̶i̶t̶t̶e̶n̶ ̶i̶n̶ ̶a̶s̶ ̶a̶ ̶g̶e̶n̶e̶r̶a̶l̶i̶s̶e̶d̶ ̶R̶i̶c̶a̶t̶t̶i̶ ̶e̶q̶u̶a̶t̶i̶o̶n̶,̶ ̶t̶h̶a̶t̶ ̶i̶s̶ ̶ $$u' = P(x) + Q(x)u + R(x)u^2, \qquad \qquad \qquad(2)$$

w̶h̶e̶r̶e̶ ̶$̶P̶,̶ ̶Q̶,̶ ̶R̶$̶ ̶a̶r̶e̶ ̶n̶o̶n̶-̶z̶e̶r̶o̶ ̶m̶e̶r̶o̶m̶o̶r̶p̶h̶i̶c̶ ̶f̶u̶n̶c̶t̶i̶o̶n̶s̶?̶ ̶

Edit: M̶y̶ ̶q̶u̶e̶s̶t̶i̶o̶n̶ ̶s̶t̶e̶m̶s̶ ̶f̶r̶o̶m̶ ̶[̶t̶h̶i̶s̶ ̶p̶a̶p̶e̶r̶]̶[̶1̶]̶,̶ ̶w̶h̶e̶r̶e̶ ̶t̶h̶e̶y̶ ̶p̶u̶t̶ ̶f̶o̶r̶w̶a̶r̶d̶ ̶a̶ ̶t̶r̶a̶n̶s̶f̶o̶r̶m̶a̶t̶i̶o̶n̶ ̶f̶r̶o̶m̶ ̶(̶2̶)̶ ̶t̶o̶ ̶(̶1̶)̶.̶ ̶I̶'̶m̶ ̶j̶u̶s̶t̶ ̶s̶t̶r̶u̶g̶g̶l̶i̶n̶g̶ ̶a̶ ̶b̶i̶t̶ ̶w̶i̶t̶h̶ ̶f̶i̶n̶d̶i̶n̶g̶ ̶a̶ ̶t̶r̶a̶n̶s̶f̶o̶r̶m̶a̶t̶i̶o̶n̶ ̶t̶h̶a̶t̶ ̶t̶a̶k̶e̶s̶ ̶(̶1̶)̶ ̶t̶o̶ ̶(̶2̶)̶.̶

Edit 2: Ignore the previous remarks. I am currently looking for a transformation that will take a general second order equation of the form

$$a(x) \frac{d^2y}{dx^2}+b(x)\frac{dy}{dx}+c(x)y=0$$

where $a(x),b(x),c(x)$ are complex-valued, meromorphic functions into the form of $(2)$.

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  • $\begingroup$ I don't understand your question. Setting $u = y$ with $P = S$, $Q \equiv 0$, and $R \equiv -1$ you have what you desire. What am I missing? $\endgroup$ Commented Aug 24, 2017 at 20:25
  • $\begingroup$ @WillieWong sorry, my mistake. I'll edit the question details now. $\endgroup$
    – user119264
    Commented Aug 24, 2017 at 20:26
  • $\begingroup$ The paper that you mentioned wants the functions to be (a) defined on the complex plane (b) meromorphic. You want functions on $\mathbb{R}$ that are smooth. Why do you think you can get an answer? $\endgroup$ Commented Aug 24, 2017 at 20:57
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    $\begingroup$ Both the forward and reverse transformation between the Riccati and a second order linear equation can be found on the corresponding Wikipedia page. $\endgroup$ Commented Aug 24, 2017 at 22:47
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    $\begingroup$ I'm voting to close this question as off-topic because the wikipedia page for Ricatti equation has the answer, and it is covered in all ode textbooks. $\endgroup$
    – Ben McKay
    Commented Aug 25, 2017 at 6:27

1 Answer 1

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The transformation $y = -R(x) + s(x) u$ takes (1) to (2) with $$\eqalign{P(x) &= {\frac { s \left( x \right)^{2}+s' \left( x \right) -S \left( x \right) }{R \left( x \right) }} \cr Q(x) &= -2 s(x)-\frac{R'(x)}{R(x)}}$$

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