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I found this question in a differential equation textbook as a question

The equation $$ \frac{dy}{dx} =A(x)y^2 + B(x)y +C(x) $$ is called Riccati's equation

show that if $f$ is any solution of the equation, then the transformation $$ y = f + \frac{1}{v} $$ reduces it to a linear equation in $v$.

I am not understanding this, what does $f$ mean? How can we find the solution with the help of the solution itself. I hope anyone could help me to solve this differential equation.

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We are given the Riccati equation:

$$\tag 1 \dfrac{dy}{dx} =A(x)y^2 + B(x)y + C(x) = A y^2 + B y + C$$

I do not want to carry around the fact that $A, B, C$ are functions of $x$.

We are asked show show that if $f$ is any solution of equation $(1)$, then the transformation:

$$\tag 2 y = f + \dfrac{1}{v}$$

reduces it to a linear equation in $v$.

First, note that they are telling us that $f$ is a particular solution to $(1)$, so just substitute $f$ into $(1)$, yielding:

$$\tag 3 f' = A f^2 + B f + C$$

But we are given the transformation $(2)$, so lets use it, we have $ y = f + \dfrac{1}{v}$, so

  • $\tag 4 y' = f' -\dfrac{1}{v^2}v' = (A f^2 + B f + C) -\dfrac{1}{v^2}v'$

But from $(1)$, we have:

  • $\tag 5 y' = A y^2 + B y + C = A\left(f + \dfrac{1}{v}\right)^2 + B\left(f + \dfrac{1}{v}\right) + C$

Equating $(4)$ and $(5)$ and collecting/cancelling like terms, leaves us with:

$\tag 6 -\dfrac{1}{v^2}v' = A\dfrac{1}{v^2} + 2 f A\dfrac{1}{v}+ B \dfrac{1}{v}$

Simplifying $(6)$, yields:

$$v' +(B + 2 fA)v = -A$$

As you can see, this has now been reduced to a linear equation in $v$, as desired.

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    $\begingroup$ I understand the proof. I just don't understand the viewpoint and mindset one needs for having such an approach. What I mean is that such an approach just doesn't seem intuitive in any sense. In the end, doing it the way Euler did it, it seemingly gets solved in a neat serendipitous way. But my question is: how did he think of coming up with such an approach? $\endgroup$ – RedHelmet Aug 31 '16 at 14:38
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    $\begingroup$ What if he thought that from a "cooked" example, knowing the solution he wanted, he built the equation and then showed the generality of doing the inverse procedure. $\endgroup$ – Alejandro Sazo Apr 9 '17 at 23:22
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The question is wrongly posed, which is probably why you don't comprehend it. Here's the correct one:

Let $y$ and $f$ be solutions to the above diff. equation such that $y=f+1/v$ for some function $v(x)$. Show that $v$ satisfies a linear diff. equation.

The solution is provided by Amzoti.

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