1. ## differential

f'(x)=1/2f(x)-4
Find f(x).

It is easy to estimate f(x).
It must be squarerootx+2. I found the result by experience. What operations can I do?

2. ## Re: differential

This is known as a differential equation. $\dfrac{d}{dx}\left(\sqrt{x}+2\right) = \dfrac{1}{2\sqrt{x}}$, so this does NOT satisfy your equation.

Instead, you wind up with $f(x) = c_1 e^{x/2}+8$ where $c_1$ is any arbitrary constant.

3. ## Re: differential

f'(x)=1/(2f(x)-4) (maybe if I write like this it will be better.)
Find f(x).

Answer is :square root2+2. But I don't know a way to solve this equation. I just made an estimation.

4. ## Re: differential

Originally Posted by kastamonu
f'(x)=1/(2f(x)-4) (maybe if I write like this it will be better.)
Find f(x).

Answer is :square root2+2. But I don't know a way to solve this equation. I just made an estimation.
The correct answer is $f(x) = c_1e^{x/2}+8$. I just told you that. $\sqrt{2}+2$ is a constant. If $f(x) = \sqrt{2}+2$, then $f'(x) = 0$, and $0 \neq \dfrac{1}{2}(\sqrt{2}+2)-4$, so you are still wrong.

5. ## Re: differential

squarerootx+2

6. ## Re: differential

Slip - you're misreading the problem. You are solving this:

$\displaystyle f'(x) = \frac 1 2 (f(x)-4)$

But the OP meant this:

$\displaystyle f'(x) = \frac 1 {2 f(x)-4}$

Starting with:

$\displaystyle \frac {dy}{dx} = \frac 1 {2y -4}$

Rearrange and you get to:

$\displaystyle (2y-4)dy = dx$

Integrate:

$\displaystyle y^2-4y-(x+c) = 0$

where c is any arbitrary constant. This then yields:

$\displaystyle y = 2 \pm \sqrt{x+c}$ .

So yes, $\displaystyle 2 + \sqrt x$ is a solution, but so is $\displaystyle 2 - \sqrt {x}$ , as well as $\displaystyle 2 \pm \sqrt {x+5}$, etc.

7. ## Re: differential

Originally Posted by kastamonu
(maybe if I write like this it will be better.)
It would be better if you learn LaTeX.

8. ## Re: differential

Many Thanks.
What is La Tex

9. ## Re: differential

Originally Posted by kastamonu
Many Thanks.
What is La Tex
LaTex Tutorial

It is how we are formatting math equations to be readable on these forums.