Results 1 to 4 of 4

Math Help - Show that the Riccati equation boils down to linear form in v

  1. #1
    Member
    Joined
    May 2009
    Posts
    211

    Show that the Riccati equation boils down to linear form in v

    I've been trying this problem but in both ways that i try it i get stuck. i would like to get any hints onto how to start it.

    An equation: dy/dx = P(x)y^2 + Q(x)y + R(x) is called the riccati form...

    If one solution, say u(x), of this is known, show that the substitution y=u+(1/v) reduces the riccati equation into a linear equation in v.

    Thnak u very much!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Well, what have you tried so far?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2009
    Posts
    211
    First I start with y=u+(1/v) and from this I get: dy/dx = u' - (1/v^2)(dv/dx)

    Then I put these last expression into the given riccati form.

    From here I tried 2 things:
    1) In the riccati form I substituded y by u(x)... this led me to call P(x)u^2 + Q(x)u + R(x) = -F(x) ... and u' = G(x)
    From here I have dv/dx + v^2G(x) = v^2F(x) ... The problem is that this is not linear, and not even bernuolli seems to be available.

    2) I substitute y=u+(1/v) in the riccati form, but it gets super ugly and it doesnt seem to go anywhere.

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Hmm. You have y'=P(x)\,y^{2}+Q(x)\,y+R(x), with u a known particular solution. Your goal is to show that the substitution y=u+1/v renders the equation linear in v. Your derivative is correct. Now plug everything into the DE:

    u'-\dfrac{1}{v^{2}}\,\dfrac{dv}{dx}=P(x)\left(u^{2}+\  dfrac{2u}{v}+\dfrac{1}{v^{2}}\right)+Q(x)\left(u+\  dfrac{1}{v}\right)+R(x).

    Rearranging a bit yields

    \underbrace{u'-P(x)u^{2}-Q(x)u-R(x)}_{\text{=0, since u is a particular solution.}}=\dfrac{1}{v^{2}}\,\dfrac{dv}{dx}+P(x)\  left(\dfrac{2u}{v}+\dfrac{1}{v^{2}}\right)+Q(x)\df  rac{1}{v}.

    Can you finish from here?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Riccati's Equation
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: November 20th 2011, 07:00 AM
  2. Riccati differential equation in Matlab
    Posted in the Math Software Forum
    Replies: 1
    Last Post: May 27th 2011, 01:16 AM
  3. Riccati Equation: existence/uniqueness relation
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: April 12th 2009, 01:58 PM
  4. Riccati Equation
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: April 3rd 2009, 12:19 PM
  5. help with riccati equation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 17th 2009, 04:00 PM

Search Tags


/mathhelpforum @mathhelpforum