1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Ricatti Equation

  1. Mar 2, 2005 #1
    Having problem with this PLEASE HELP ME!

    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.
    Go to Top of Page
  2. jcsd
  3. Mar 2, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper

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

    How about posting some of your work??

  4. Mar 2, 2005 #3
    I guess it could also have been:

    [tex]\frac{dy}{dx}=-y+a(x)y^{2}+b(x) [/tex]

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

    [tex]\frac{dy}{dx}=-y^{2}+a(x)y+b(x) [/tex]

    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.

  5. Mar 2, 2005 #4


    User Avatar
    Science Advisor
    Homework Helper

    Please, first thing first: The Riccati equation has [itex] y^2 [/itex] as Daniel pointed out.

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


    Yoyo, can you report the answer to the first question, i.e., substitute [itex]y=\frac{u^{'}}{u}[/itex] 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.

  6. Mar 6, 2005 #5


    User Avatar
    Science Advisor
    Homework Helper

    A summary



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

    [tex]u^{''}-au^{'}-bu=0 [/tex]

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


    Separating variables gives:


    Factoring via partial-fraction decomposition, we get:


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


    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 [itex]y_0[/itex]), so that:


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




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

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


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


    Multiplying top and bottom by [itex]e^x[/itex] yields:


    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?

    Attached Files:

    Last edited: Mar 6, 2005
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook