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Consider the differential equation $$f'(x)=\frac{f(x)}{2x} - \sqrt{f(x)}. $$

Per Wolfram Alpha, its solution is $$f(x)=\frac{1}{9}(4x^2-12cx^{5/4}+9c^2 x^{1/2}).$$

From the differential equation, we can see that when $x$ and $f(x)$ are large, the function has a negative slope. However, we see from the solution that when $x$ is large, so is $f(x)$, yet the function has a positive slope. What explains this apparent contradiction in intuition?

Here is an example with concrete values. Setting $c=1$, $f(20) = 125$. From the solution, $f$ slopes upwards at $x=20$. Yet according to the differential equation, $f'(20) = 125/40-125^{0.5} = -8$, which suggests $f$ slopes downwards. What explains this disconnect?

Here is an interactive Desmos graph showing these functions and calculations. As you can see, the derivative of $f$ as calculated by Desmos is markedly different from the slope given by the differential equation. Is there any chance Wolfram Alpha got this wrong? Or is there some other explanation for the seeming contradiction?

Any feedback or guidance would be greatly appreciated.

user2813920183
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    Could you explain more why when $x$ and $f(x) $ are large the slope should be negative? – Jonatan Garcia Jul 15 '21 at 03:16
  • @JonatanGarcia, thank you. For instance, setting $c=1$, $f(20) = 125$. From the solution, $f$ slopes upwards at $x=20$. Yet according to the differential equation, $f'(20) = 125/40-125^{0.5} = -8$, which suggests $f$ slopes downwards. – user2813920183 Jul 15 '21 at 03:23
  • Well, that means the function is decreasing, the function can be always positive and decreasing, for example $e^{-x}$ is that your question? – Jonatan Garcia Jul 15 '21 at 03:26
  • @JonatanGarcia, thank you very much but I am not sure I agree. $f(20)$ is both positive and increasing, yet the derivative according to the diff eq is negative. – user2813920183 Jul 15 '21 at 03:28
  • Plot the graph on Desmos or Geogebra, I don't see why the function is in increasing – Jonatan Garcia Jul 15 '21 at 03:44
  • @JonatanGarcia I linked to the Demos plot in the question, it does look to be increasing. – user2813920183 Jul 15 '21 at 03:45
  • Ok, I see, hmm, maybe the found solution satisfies the differential equation only in one interval, you can see in the Desmos that the derivative of the solution and the derivative given by the differential equation are equal near the origin, depending of the value of $c$. – Jonatan Garcia Jul 15 '21 at 03:57
  • This is Bernoulli equation with solution $\sqrt f=C\sqrt[4]x-\frac{2x}{3}$. – Aleksas Domarkas Jul 15 '21 at 05:51
  • Taking an initial condition $f(x_0) = f_0$ gives a quadratic equation on $c$, we cannot choose a root arbitrarily. A solution corresponding to the other root cannot be given by the W|A formula at $x = x_0$. Suppose we take $f(1) = 1/9$, which corresponds to $c = 1$, then the solution will be $$f(x) = \cases { (4 x^2 - 12 x^{5/4} + 9 \sqrt x)/9 & $1 \leq x \leq x_1$ \ 0 & $x_1 < x$},$$ where $x_1 = (3/2)^{4/3}$. Because of $\sqrt f$, a solution near $(x, f) = (x_1, 0)$ is not unique. – Maxim Sep 09 '21 at 19:06

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Ignoring the $f=0$ solution, we can rearrange the equation to see that

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

a linear equation in $\sqrt{f}$, which we can solve by integrating factor

$$\int\left(\frac{\sqrt{f(x)}}{x^{\frac{1}{4}}}\right)' = \int-\frac{1}{2x^{\frac{1}{4}}} \implies \sqrt{f(x)} = Cx^{\frac{1}{4}}-\frac{2}{3}x$$

However, at this stage, $C$ is not allowed to be any real number - it is dependent on the domain of the function. For positive $x$ we have that

$$Cx^{\frac{1}{4}}-\frac{2}{3}x\geq 0 \implies C \geq \frac{2}{3}x^{\frac{3}{4}}$$

Your choice of $C=1$ was invalid because extending the domain of the function to $20$ puts a lower bound on $C$ to be

$$C \geq \frac{2}{3}(20)^{\frac{3}{4}} \approx 6.3049$$

Alternatively, once $C$ is chosen, it provides an upper bound to the domain of the function, in which case $C=1$ puts the maximal extent of the interval of existence to be

$$0 < x \leq \left(\frac{3}{2}C\right)^{\frac{4}{3}} = \left(\frac{3}{2}\right)^{\frac{4}{3}} \approx 1.717$$

You cannot have arbitrarily large $x$.

Ninad Munshi
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