1. ## Small question about Riccati

The form of Riccati is $y' = P(x)y^{2} + Q(x)y + R(x)$
the book says that $y' = \frac{1}{x^{2}}y^{2} - \frac{1}{x}y + 1$ is a Riccati equation.
So how does the constant 1 qualify as a R(x) term?

2. ## Re: Small question about Riccati

Originally Posted by catenary
The form of Riccati is $y' = P(x)y^{2} + Q(x)y + R(x)$
the book says that $y' = \frac{1}{x^{2}}y^{2} - \frac{1}{x}y + 1$ is a Riccati equation.
So how does the constant 1 qualify as a R(x) term?
1 is a perfectly legitimate function of x.

$R(x)=1$

3. ## Re: Small question about Riccati

is it a legitimate function of x because you can write it as $1 = 1(x)^{0}$ ?

4. ## Re: Small question about Riccati

Originally Posted by catenary
is it a legitimate function of x because you can write it as $1 = 1(x)^{0}$ ?
sure if you like but even if you couldn't it's still a function of x. It just happens to be a constant function.

Does it map every value in it's domain to a unique number? Yes. It maps every value in it's domain to 1. That's all you need to be a function.