# Math Help - Riccati's Equation

1. ## Riccati's Equation

Greetings,
I came across the folllowing question on Ross, Differential Equations and was unable to solve it:
$y'=-y^2+xy+1$ with one solution given as $f(x)=x$, you are asked to find the general solution, my solution is as follows:
$y=f(x)+\frac{1}{v}$
$\frac{d}{dx}(x+\frac{1}{v})=-(x+\frac{1}{v})^2+x(x+\frac{1}{v})+1$
$-\frac{1}{v^2}\frac{dv}{dx}=-\frac{x}{v}-\frac{1}{v^2}$
Writing it in the standard form leads:
$\frac{dv}{dx}-vx=1$
I found the integrating factor:
$e^{\int{-xdx}}=e^\frac{-x^2}{2}$
My final result is:
$v=e^\frac{x^2}{2}\int{e^\frac{-x^2}{2}dx}+ce^\frac{x^2}{2}$
Afterwards I could not evaluate the indefinite integral and this seems to be an introductory question, do I have a mistake here ?

Any help is appreciated.
Thanks

2. ## Re: Riccati's Equation

Looks good to me!

3. ## Re: Riccati's Equation

I did a quick search on Wolfram Alpha and I think that integral cannot be evaluated with the use of elementary functions and it is called the error function if I am not mistaken?