# Thread: Simple differential equation problem

1. ## 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?

2. Originally Posted by chengbin
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?)

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

3. 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?

4. 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.

5. 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 $\displaystyle x+y=u$