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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.
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  2. jcsd
  3. Mar 2, 2005 #2


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    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


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    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


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    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
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