How to solve the Riccati's differential equation
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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$\begingroup$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.
$\endgroup$ 2 $\begingroup$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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