Results 1 to 5 of 5

Math Help - Simple differential equation problem

  1. #1
    Senior Member
    Joined
    Jan 2009
    Posts
    290

    Simple differential equation problem

    I don't want to type this out, so I took a picture of it.





    I don't understand the beginning differentiating part. [Why f(x) x f'(y)]

    How does letting y=0 give you f'(y)=A (I know A is just a variable, using that as a value of something)

    How does f'(x)=Af(x)?

    I don't get the substituting x=y=0 part on the bottom, why x=y=0?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,406
    Thanks
    1294
    Quote Originally Posted by chengbin View Post
    I don't want to type this out, so I took a picture of it.





    I don't understand the beginning differentiating part. [Why f(x) x f'(y)]

    How does letting y=0 give you f'(y)=A (I know A is just a variable, using that as a value of something)

    How does f'(x)=Af(x)?

    I don't get the substituting x=y=0 part on the bottom, why x=y=0?
    This is an application of the product rule. Their notation is a little off seeing as when you differentiate with respect to y, it is a PARTIAL derivative.

    If f(x + y) = f(x)\cdot f(y)

    Then \frac{\partial}{\partial y}[f(x + y)] = \frac{\partial}{\partial y}[f(x) \cdot f(y)]

     = f(x) \frac{\partial}{\partial y}[f(y)] + f(y) \frac{\partial}{\partial y}[f(x)]

     = f(x) \frac{\partial f(y)}{\partial y} + 0\cdot f(y). (Why?)

     = f(x) \frac{\partial f(y)}{\partial y}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,406
    Thanks
    1294
    As for the
    Why substitute y = 0?
    assuming their logic and notation is correct (I have my suspicions...) notice that you are trying to find

    f'(x).

    But we are not given f'(x), it's f'(x + y).

    But if y = 0, then

    f'(x + y) = f'(x + 0) = f'(x), which is what we're trying to find.

    Remember that f'(x + y) = f(x) \cdot f'(y), so

    f'(x) = f'(x + 0) = f(x) \cdot f'(0).



    For any function of y, this function depends ONLY on y. Substituting ANY value for y ALWAYS gives you a constant.

    f'(y) is one such function of y.

    Since 0 is a constant, f'(0) is also a constant.

    Thus f'(0) = A, where A is a constant.


    Therefore f'(x) = f(x) \cdot A

    f'(0) = Af(x).


    Is that a bit clearer?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,347
    Thanks
    30
    Here's another way maybe a little clearer. Assume that F \ne 0 clearly it's true if F = 0. Take the natural log's of both sides

    \ln f(x+y) = \ln f(x) + \ln f(y).

    Now differentiate wrt to x and then y. So

    \left(\frac{f'(x+y)}{f(x+y)}\right)' = 0.


    Then set either  x \, \text{or}\, y = 0 . You now have an ODE which can be integrated easily. Pretty much like the later part of the book.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Jan 2009
    Posts
    290
    Sorry for bothering you guys with these noob questions.

    I have another one, and this is probably a very dumb question.



    Where did the 1 come from in the beginning when they differentiate both sides?

    nvm. Damn, that's a noob question. I thought they're differentiating \frac {dy} {dx}=x+y. They're actually differentiating x+y=u
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Simple Differential Equation
    Posted in the Differential Equations Forum
    Replies: 6
    Last Post: March 21st 2011, 06:40 PM
  2. probably very simple differential equation :D
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: September 12th 2010, 10:09 PM
  3. Replies: 2
    Last Post: March 15th 2010, 03:33 AM
  4. Simple differential equation problem
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: April 6th 2009, 04:38 PM
  5. simple differential equation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 18th 2007, 12:13 PM

Search Tags


/mathhelpforum @mathhelpforum