# Riccati Equation

• Apr 2nd 2009, 12:02 PM
tn2k7
Riccati Equation
Consider the initial value problem for a Riccati equation:

$y' + y^{2} + 1 = 0, y(0) = 1$

Find the true solution y(x). How many solutions have you found?

This is actually part of a project involving us researching the riccati equation. So, I haven't learned anything about this topic in class, so I'm really not sure how to begin.

I've found guide on the website http://www.sosmath.com/diffeq/first/...i/riccati.html but I'm not sure how reliable it is.
• Apr 3rd 2009, 12:09 AM
chisigma
Only for clarification purpose: the 'initial value problem' to be 'attacked' is this...

$y^{'}= y^{2} + y + 1$ , $y(0)=1$

... or some different?...

Kind regards

$\chi$ $\sigma$
• Apr 3rd 2009, 08:28 AM
tn2k7
Sorry, I copy and pasted and the subscripts and superscripts came out wrong. It should be:

$y' + y^{2} + 1 = 0, y(0) = 1$
• Apr 3rd 2009, 01:19 PM
chisigma
All right!... the Riccati equation is written as...

$y^{'}= p(x) + q(x)\cdot y + r(x)\cdot y^{2}$ (1)

In general the solution of (1) is possible only if you know one particular solution and that is in most cases an hard problem. The equation you have proposed however is...

$y^{'}= -(1+y^{2})$ , $y(0)=1$ (2)

... and, because the independent variable x doesn't partecipate, we can use a different approach. Let's suppose in (2) to swap the x and y, so that the equation becomes...

$x^{'}= -\frac{1}{1+y^{2}}$ , $x(1)=0$ (3)

In this case you can separate the variables, integrate immediately and obtain...

$x= - \tan^{-1} y + c$ (4)

The constant c can be obained with the condition $x(1)=0$ and that gives $c= - \frac {\pi}{4}$. The (4) is easily invertible, so that the 'true solution' of (2) is...

$y= -\tan (x - \frac{\pi}{4})$ (5)

Kind regards

$\chi$ $\sigma$