Your differential system looks like this: $$\dot x=Ax\qquad x(0)=x_0$$ where $A$ is a $2\times 2$ matrix. Suppose that $x_0$ is an eigenvector of $A$. Then, when $t=0$, $\dot x=\lambda x_0$, so the derivative of the solution curve points in the direction of $x_0$. If $\lambda<0$, the curve is moving towards the origin. If $\lambda>0$, the curve is moving away from the origin.
These eigendirections give us a "skeleton" around which to build the geometry of the set of solutions to the differential system. If we have two different eigenvectors $v_1$ and $v_2$ with eigenvalues $\lambda_1$ and $\lambda_2$, (which your question implicitly assumes), then we can write any $x(t)$ as $$x(t)=a(t)v_1+b(t) v_2$$ for some scalar functions $a$ and $b$. Then $$\dot x=\dot a(t)v_1+\dot b(t) v_2=Ax=a(t)Av_1+b(t)Av_2=\lambda_1 a(t)v_1+\lambda_2 b(t) v_2$$ which, by the uniqueness of the decompositions in the base $(v_1,v_2)$, implies that $$\dot a(t)=\lambda_1a(t)\qquad \dot b(t)=\lambda_2b(t)$$ So the behavior of the solutions of the system is really dependent on the behavior along each eigenvector.
- If $\lambda_1<0<\lambda_2$, then one component, or one part of the solution is getting pulled toward the origin, while the other part is repelled. This leads to the saddle.
- If $\lambda_1< \lambda_2<0$, then both parts of the solution are pulled toward the origin, and we get a sink.
- If $0<\lambda_1< \lambda_2$, then both parts are pushed away from the origin, and so we get a source.
As for why we can tell all of this without knowing the specifics of the matrix, the answer is that if a matrix has two distinct eigenvalues, we know everything about the behavior of that matrix just from its eigenvalues and eigenvectors.
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 $x$.
Best Answer
Very good intuition. The only flaw in your thinking is that in the original equation, $\frac{dy}{dx}$ is multiplied by $x$.
Therefore, when $x = 0$, it becomes irrelevant what $\frac{dy}{dx}$ is.
Addendum Reaction to OP's comment/reaction re 10-3-2020.
The original equation is $x\frac{dy}{dx} - y = x^3.$
You discovered that the solution is not one equation but a family of equations, represented by
$y = \frac{x^3}{2} + cx ~\Rightarrow ~\frac{dy}{dx} = f'(x) = \frac{3x^2}{2} + c.$
Your intuition then rebelled, intuiting (in effect):
Something is wrong here.
Consider two separate solutions:
$f_1(x) = \frac{x^3}{2} + c_1x.$
$f_2(x) = \frac{x^3}{2} + c_2x ~: ~c_2 \neq c_1$
As the value of $c$ changes, so does the value of $\frac{dy}{dx}.$
This means, that at any given value of $x$, $f'_1(x)$ and $f'_2(x)$ will be unequal.
How can a family of functions, as represented by $f_1(x)$ and $f_2(x)$
each of whom must have a different value for $f'(x_0)$ at one specific value of $x_0$ ever intersect?
Answer:
My original answer gave only a mathematical explanation for why
(for example) $f_1(x)$ and $f_2(x)$ can intersect at the origin
despite the fact that $f'_1(0) \neq f'_2(0).$
Intuitively, consider the alternative set of two differential equations:
$f''(x) = 0,$ for all $x~~$ combined with $~~f(x) = 0.$
The above two equations will be satisfied by the family of equations:
$f(x) = cx,~$ all of which intersect at $x=0$.
Just because each member of the family has a different derivative at $x=0$
doesn't mean that they can't all intersect at $x=0$.
As far as the specific terminology in your comment, although I have had a brief exposure to the concepts of isocline and directional field, I lack the professional experience to grapple with these concepts and explain the supposed flaw in your intuition (if any).
It seems to me that you are asking a legitimate question that is simply beyond my expertise. If I were you, and I couldn't come to terms with your pending question in 24 hours, and if no one else responded because mathSE queries tend to get lost in the shuffle, then I would:
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This way you will (in effect) be (quite reasonably) attempting to force qualified mathSE reviewers to focus on the issue that your perspective deems as pending.
(2)
In this query, leave your comment just where it is. However, add an addendum to your original query in this posting. In this addendum, repeat the pending question of your pertinent comment. Indicate that from your perspective, this is a pending question. Indicate (also in the addendum) that you are construing the pending question to be a second question, and you have therefore initiated a 2nd mathSE query. Provide a link to the 2nd mathSE query in the addendum of this original query.