I am trying to solve a simple Ricatti equation and am not sure i'm correct. Since this is probably the simplest common case I thought I'd post it here:

Write the Riccati equation as \begin{equation*} \frac{d E (\tau) }{d \tau} = e_2 E(\tau)^2 + e_1 E(\tau) + e_0 \end{equation*} where we can reduce to a second order linear equation via the substitution $E(\tau) = - \frac{1}{e_2} \frac{w^\prime (\tau)}{w(\tau)}$ which yields \begin{equation*} w^{\prime \prime} - e_1 w^\prime + e_2 e_0 w = 0. \end{equation*} The general solution to this ODE is $w(\tau) = C_{1} e^{r_{1} \tau} + C_{2} e^{r_{2} \tau}$ where $r_1$ and $r_2$ are the positive and negative roots \begin{align*} r_{1,2} = \frac{e_1 \pm \sqrt{e_1^2 - 4 e_2 e_0}}{2} = \frac{e_1 \pm q}{2} \end{align*} and $C_{1,2}$ are constants determined by the boundary condition. A general solution for $E(\tau)$ is then \begin{align*} E(\tau) &= - \frac{1}{e_2} \frac{C_1 r_{1} e^{r_{1} \tau} + C_2 r_{2} e^{r_{2} \tau}}{C_1 e^{r_{1} \tau} + C_2 e^{r_{2} \tau}} = - \frac{1}{e_2} \frac{C r_{1} e^{r_{1} \tau} + r_{2} e^{r_{2} \tau}}{C e^{r_{1} \tau} + e^{r_{2} \tau}} \end{align*}

Does this look correct? How else can these ODEs be solved?

  • 1
    $\begingroup$ Am I missing the question? $\endgroup$ – Amzoti May 27 '13 at 21:24
  • $\begingroup$ yes - sorry! edited. $\endgroup$ – Luap Nalehw May 27 '13 at 21:57
  • $\begingroup$ If your $e_0,\,e_1,\,e_2$ are constants, then you equation is separable and can be integrated directly. $\endgroup$ – Artem May 28 '13 at 8:17

I suppose that your non linear OE $$\frac{d E (\tau) }{d \tau} = e_2 E(\tau)^2 + e_1 E(\tau) + e_0 $$ is rewritten as $$\frac{d E }{d x} = R(x) E^2 + Q(x)E + P(x),~~~(*)$$ just for the simplicity in indexes and the variable $\tau$ is our usual variable $x$. Another method in which you can overcome the OE would be to use a known particular solution of $(*)$, say $y_1$. Then set $y=y_1+u$ wherein $u$ is a solution of $$\frac{du}{dx}-(Q+2y_1R)u=Ru^2$$ Since the latter OE is a Bernoulli equation with $n=2$, it can be turned into a linear OE: $$\frac{dw}{dx}+(Q+2y_1R)w=-R$$ where $w=u^{1-2}=u^{-1}$.

  • $\begingroup$ @flashdesign2550: Thanks. :) but let us to be seen by the OP. $\endgroup$ – mrs May 28 '13 at 8:38
  • $\begingroup$ Thanks. So three methods: (1) Direct integration (2) Reduction to second order linear ODE (3) Reduction to Bernoulli equation $\endgroup$ – Luap Nalehw May 28 '13 at 14:09

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