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?
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.
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