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Given a Riccati Equation which is differential equation of the form:

$$ \frac{dy}{dx} = a_0 (x) + a_1 (x)y + a_2 (x)y^2 $$

It is well known that the transformation:

$$ y = -\frac{1}{a_2(x)} \frac{\frac{du}{dx}}{u} $$

Can be substituted into this equation to transform it into a second order linear differential equation (which after simplifying the algebra is)

$$ a_0(x)a_2(x) u - (a_1(x)a_2(x) + 1) \frac{du}{dx} + a_2(x)\frac{d^2u}{du^2} = 0 $$

My question is: How does one derive this transformation from scratch. Since although it is easy to prove that the transformation works it does not reveal how it was found.

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The important point is not that the transformed equation is of second order, but that it is linear. Often, a linear second order ODE is easier to solve than a non linear first order ODE.

If a particular solution of the Riccati equation is known, there is no interest to transform the Riccati equation, because it could be directly solved. But it is not always possible to first find a particular solution. Then, the transformation to a linear second order ODE can be helpfull in many casses, especially when the solutions involve special functions.

For example : $y' = -2\frac{y}{x}+\frac{y^2}{x}+x $ can be transformed to a second order ODE of Bessel's kind which allows to solve the Riccati equation in terms of Bessel functions. Without this transformation, it would be almost impossble to first find a particular solution an so, quite impossible to solve the non linear first order ODE.

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    $\begingroup$ Thanks for the clarification! :) and yes I specifically meant linear (will probably edit post to include that). Still this doesn't answer how the transformation is motivated. It right now appears to me that someone by coincidence found the trick. How do you prove that such a trick actually exists and then logically deduct what that 'trick is' that allows you to convert the riccati to a second order linear ode. $\endgroup$ – frogeyedpeas Jun 8 '14 at 13:35
  • $\begingroup$ I cannot see any trick. Do you think that there a trick behind any method of solving a problem ? It is strange to think that the Mathematicians discover the solutions by coincidence ! In the present case, probably an Historian could say when and how the said transformation was found and published. I think that the inverse transformation was used before, in trying to transform a second order ODE to a first order one, but unfortunately non linear. It's not much genious to do it in the opposite direction when there is an advantage to do so. $\endgroup$ – JJacquelin Jun 8 '14 at 14:12
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    $\begingroup$ Allow me to rephrase: So given a riccati equation. You ask the question 'Can this be transformed to a second order linear ODE'. How do I necessarily answer yes/no to that (without having to find such a transformation and then furthermore: if the answer is yes, how do I derive a substitution to convert the Riccati to a second order linear ODE (for example, consider how the derivation of the quadratic formula is done by perform standard 'completing the square' technique on a general quadratic). $\endgroup$ – frogeyedpeas Jun 8 '14 at 14:34
  • $\begingroup$ Sorry, I cannot fully understand you want and what is exactly the meaning of your question, which wording had changed a lot. I would not go, by missunderstanding, into a fishy discussion. I hoppe that someone more aware than me about these kind of questions will help you. $\endgroup$ – JJacquelin Jun 8 '14 at 15:07
  • $\begingroup$ Alright then, thanks anyways for attempting to help :) $\endgroup$ – frogeyedpeas Jun 8 '14 at 20:27
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The things look more natural if you start with second order ODE $$(rx')'+cx=0$$ and use the transformation $w=\frac{rx'}{x}$ to get Riccati type equation $$w'+c+r^{-1}w^2=0.$$

The numerator of $w$ is just the quasiderivative $rx'$ and the reason for $x$ in denominator could be the homogeneity. Not sure who was the first, but a lot of info can be found in this classical book by Swanson: Comparison and oscillation theory of linear differential equations.

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