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?

  • 1
    $\begingroup$ 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.] $\endgroup$
    – Pacciu
    Dec 22, 2011 at 15:30

1 Answer 1


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.


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