Ricatti Equation

1. Mar 2, 2005

yoyo

Consider the following Riccatti equation:

dy/dx= -y+ a(x)y + b(x) (Eq. 2)

Here a(x) and b(x) are arbitrary functions.

1. Set y(x)= u'(x)/ u(x) where u(x) is a function to be determined. Use (Eq. 2) to show that u(x) satisfies a linear differential equation of second order.

2. Set a=0 and b=1. Solve (Eq. 2) by separating variables using y(0)=0 as the initial condition.

3. Find the function u(x) corresponding to the case a=0, b=1 and y(0)=0. When solving the second order equation for u(x) assume that u'(0)= y(0) and u(0)= 1. Compare the solution of (Eq. 2) from Part 2 with u'(x)/u(x).

4. Eplain why it is nice to replace a non-linear first order differential equation with a linear second order one.
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2. Mar 2, 2005

dextercioby

It looks very simple.I think you meant to write:
$$\frac{dy}{dx}=-y^{2}+a(x)y+b(x)$$ (2)...

Daniel.

3. Mar 2, 2005

Jayboy

I guess it could also have been:

$$\frac{dy}{dx}=-y+a(x)y^{2}+b(x)$$

but given the requirements of part 2, I think you are right, it should be:

$$\frac{dy}{dx}=-y^{2}+a(x)y+b(x)$$

Yoyo - the problem you have been given is very explicit about what is required. If you show a little of what you have done and where you are having difficulty it will be easier to help.

J.

4. Mar 2, 2005

saltydog

Please, first thing first: The Riccati equation has $y^2$ as Daniel pointed out.

Thus, let's assume then that yoyo made a typo and go with:

$$y^{'}=-y^2+a(x)y+b(x)$$

Yoyo, can you report the answer to the first question, i.e., substitute $y=\frac{u^{'}}{u}$ in the equation above and some things cancel out, divide by u to clean up a bit, and then end up with a second-order ODE?

If this thing dies out and nobody responds for a few days, I'm gonna' follow-up with a full report and I don't care if nobody reads it either. Whatever. You know if you jump out of a plane, you have to write a Riccati equation on you sleeve else you won't know how to fall right.

Salty

5. Mar 6, 2005

saltydog

A summary

For:

$$y^{'}=-y^2+ay+b$$

Letting $y=\frac{u^{'}}{u}$ and substituting this into the ODE, we get the converted version in terms of u(x):

$$u^{''}-au^{'}-bu=0$$

Letting a=0 and b=1 we can avoid converting to u(x) and separate variables:

$$\frac{dy}{dx}=1-y^2$$

Separating variables gives:

$$\frac{dy}{1-y^2}=dx$$

Factoring via partial-fraction decomposition, we get:

$$(\frac{1}{2(1+y)}+\frac{1}{2(1-y)})dy=dx$$

Integrate indefinitely, convert from logarithms to exponents, and keep up with the constant of integration, K, produces:

$$y(x)=\frac{Ke^{2x}-1}{1+Ke^{2x}}$$

Substituting the initial conditions y(0), we find K=1 or you could have just performed a definite integration above, (you know, from y to $y_0$), so that:

$$y(x)=\frac{e^{2x}-1}{1+e^{2x}}$$

Starting with the equation for u(x), we have:

$$u^{''}-u=0$$

or:

$$u(x)=c_1e^x+c_2e^{-x}$$

Now, the initial conditions for u(x) are:

u(0)=1 and u'(0)=0. Substituting these, we get:

$$u(x)=\frac{1}{2}e^x+\frac{1}{2}e^{-x}$$

Since $y(x)=\frac{u^{'}(x)}{u(x)}$, upon differentiating this expression, we get for y:

$$y(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}$$

Multiplying top and bottom by $e^x$ yields:

$$y(x)=\frac{e^{2x}-1}{e^{2x}+1}$$

which is the same answer as above.

A plot is attached. Notice that as x increases without bound, y approaches a limiting factor. Remember I stated that the Riccati equation is used to describe jumping from a plane? Note in free-fall, you reach a "terminal velocity". Isn't someone here working on that problem?

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Last edited: Mar 6, 2005