Results 1 to 5 of 5

Math Help - Implicit Differentiation and then Euler's method

  1. #1
    Newbie
    Joined
    May 2009
    From
    Victoria
    Posts
    16

    Implicit Differentiation and then Euler's method

    In my studies, I encountered a question which I did not know how to do. The answer referred to implicit differentiation but I did not understand the steps undertaken, so could someone explain to me how to do the following question?

    \frac{dy}{dx}=y(1-y), show that \frac{d^2y}{dx^2} = (1-2y)y(1-y).
    The answer states that \frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx}) = \frac{d(y-y^2)}{dy} \cdot y(1-y) = (1-2y)y(1-y). Where does the y - y^2 come from?


    Then I had problem with Euler's method with the question:
    Given that y=2 when x=0, use Euler's method with a step-size of \frac{1}{4} to estimate the value of y when x = \frac{1}{2}
    How do I use Euler's method if all I have if f(y)?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,653
    Thanks
    1478
    Quote Originally Posted by Enedrox View Post
    In my studies, I encountered a question which I did not know how to do. The answer referred to implicit differentiation but I did not understand the steps undertaken, so could someone explain to me how to do the following question?


    The answer states that \frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx}) = \frac{d(y-y^2)}{dy} \cdot y(1-y) = (1-2y)y(1-y). Where does the y - y^2 come from?

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


    \frac{d^2y}{dx^2} = \frac{d}{dx}[y(1 - y)]

     = y\frac{d}{dx}(1 - y) + (1 - y)\frac{d}{dx}(y), from use of the product rule...

     = y\left(-\frac{dy}{dx}\right) + (1 - y)\frac{dy}{dx}

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

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

     = y(1 - y)(1 - 2y).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,957
    Thanks
    1631
    Quote Originally Posted by Enedrox View Post
    In my studies, I encountered a question which I did not know how to do. The answer referred to implicit differentiation but I did not understand the steps undertaken, so could someone explain to me how to do the following question?


    The answer states that \frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx}) = \frac{d(y-y^2)}{dy} \cdot y(1-y) = (1-2y)y(1-y). Where does the y - y^2 come from?


    Then I had problem with Euler's method with the question:

    How do I use Euler's method if all I have if f(y)?
    I don't know what you mean by that. The problem appears to be that you don't have f(x)! You don't say how the problems are arranged on the page. Could they be referring to the previous differential equation?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    May 2009
    From
    Victoria
    Posts
    16
    lol, the solution is so simple now that I actually understand it Thank you.

    Although I would have done it slightly differently (now that I understand what's going on):
    \frac{d^2y}{dx^2} = \frac{d}{dx}[y(1 - y)]=\frac{d}{dx}[y-y^2)]
    =1 \cdot \frac{dy}{dx} - 2y\frac{dy}{dx}
    =\frac{dy}{dx}(1-2y)
    =y(1-y)(1-2y)


    As for Euler's method, it is question 4 e (section 2). from the 2006 VCE specialist math exam
    Last edited by Enedrox; October 31st 2009 at 01:05 AM. Reason: Amended comment about Euler's method
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,957
    Thanks
    1631
    So the problem is to solve \frac{dy}{dx}= y(1-y) y(0)= 2, using Euler's method with a step of 1/4, from 0 to 1/2. Did you expect us to guess that before?

    Well, that's easy enough, you only have to do two steps! Since y(0)= 2, at 0, \frac{dy}{dx}, at x= 0, is 2(1-2)= -2. The line with slope -2, starting at (0,2) is y= -2x+ 2 and at x= 1/4, y(1/4)= -2(1/4)+ 2= -1/2+ 2= 3/2.

    Now do the same thing again. If y= 3/2, what is dy/dx= y(1- y)? What is the equation of the line through (1/4, 3/2) with that slope? What is y on that line when x= 1/2?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: December 9th 2010, 12:21 AM
  2. Solve the following ODE? by dint of an implicit euler
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: September 28th 2009, 09:35 AM
  3. Euler Method
    Posted in the Math Software Forum
    Replies: 5
    Last Post: March 15th 2009, 02:58 PM
  4. Euler's Method
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: March 11th 2009, 05:33 PM
  5. Implicit Euler
    Posted in the Calculus Forum
    Replies: 0
    Last Post: February 18th 2009, 02:42 AM

Search Tags


/mathhelpforum @mathhelpforum