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Thread: Simple differential equation problem

  1. #1
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    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?
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  2. #2
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    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 $\displaystyle f(x + y) = f(x)\cdot f(y)$

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

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

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

    $\displaystyle = f(x) \frac{\partial f(y)}{\partial y}$.
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  3. #3
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    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

    $\displaystyle f'(x)$.

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

    But if $\displaystyle y = 0$, then

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

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

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



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

    $\displaystyle f'(y)$ is one such function of y.

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

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


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

    $\displaystyle f'(0) = Af(x)$.


    Is that a bit clearer?
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  4. #4
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    Here's another way maybe a little clearer. Assume that $\displaystyle F \ne 0$ clearly it's true if $\displaystyle F = 0$. Take the natural log's of both sides

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

    Now differentiate wrt to x and then y. So

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


    Then set either $\displaystyle x \, \text{or}\, y = 0 $ . You now have an ODE which can be integrated easily. Pretty much like the later part of the book.
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    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 $\displaystyle \frac {dy} {dx}=x+y$. They're actually differentiating $\displaystyle x+y=u$
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