# Relationship between solutions of second order differential equation derived from Ricatti equation

I am attempting to solve the Ricatti equation:

$$3x^2 y' - 2y^2 - x y +2=0$$

And I have attempted the method suggested on the Wikipedia page of converting it to a linear second order differential equation. I have solved this second order ODE, but this comes with two solutions. That is, calling our solution $u(x)$ I have solved it to find

$$u(x) = \alpha u_1(x) + \beta u_2(x)$$

However, when re-substituting to find the equation y(x), I will still have 2 unknown quantities (both $\alpha$ and $\beta$), with only one initial condition. It strikes me that there must be some relationship between $\alpha$ and $\beta$, as the original equation is only first order and should only require one initial conditions. However, I cannot find any literature on this.

Is my supposition correct? And if so, what is the relationship between the two constants?

• Please, check the spelling. The correct name of your ODE is Riccati equation (with two "c"s and one "t") because it's named after Italian mathematician Jacopo Francesco Riccati (1676 - 1754). [BTW, "ricatti" is the plural of the Italian word "ricatto" which translates into blackmail.] – Pacciu Dec 22 '11 at 15:30

The Riccati equation $$y'=a(x)\,y^2+b(x)\,y+c(x)$$ is transformed into the second order linear equation $$u''-\Bigl(\frac{a'}{a}+b\Bigr)u'+a\,c\,u=0$$ in the unknown $u(x)$ by the change $$y=-\frac{1}{a(x)}\,\frac{u'}{u}.$$ If the solution of this equation is $u=C_1u_1+C_2u_2$, then the solution of the original equation is $$y=-\frac{1}{a(x)}\,\frac{C_1u'_1+C_2u'_2}{C_1u_1+C_2u_2}=-\frac{1}{a(x)}\,\frac{u'_1+K\,u'_2}{u_1+K\,u_2}\ ,$$ where $K=C_2/C_1$ if $C_1\ne0$; a similar formula holds if $C_2\ne0$. So, the solution of the Riccati equation depends only on one constant.