Graphical Representation of Differential Equations This may seem like a fairly stupid question since I'm new to learning differential equations.
I'm solving the following differential equation $$x\frac {dy}{dx}-y=x^3$$ and the solution I come up with is $$y=\frac{x^3}{2}+cx$$ but what bothers me is that solution is not unique...for different values of c I could get different solutions and all those solutions intersect at the origin.
Isn't this wrong? I mean what I understood from a graphical perspective is that every point in the direction field can have only one slope. Multiple solutions with differents c's here imply multiple slopes at the origin.
 A: 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
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A: Since you have not recited a theorem which will support your claim that there cannot be a point common to two otherwise distinct solutions, I'll do that for you.  From M.I.T. 18.03 Ordinary Differential Equations: Notes and Exercises, $\S$G p. 2 (PDF p. 7), the Intersection Principle

Integral curves of $y' = f(x,y)$ cannot intersect wherever $f(x,y)$ is smooth.

So, first, you need to state your equation in the form in the theorem.
$$  \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{y}{x} + x^2  $$
The right-hand side ("$f(x,y)$" in the theorem) is not smooth at $x = 0$ because $\frac{y}{x}$ is undefined at $x = 0$.  The theoreom promises integral curves do not intersect for all $x \neq 0$, but the theorem makes no promises when $x = 0$.  You have found an example for why it can make no promises at $x = 0$.
A: The equation of $f'(x)$ (or $y'$) you get when you differentiate is the equation of the tangent of $f(x)$. For a polynomial of degree $n$, the tangent is a polynomial of degree $n-1$. The slope of $f(x)$ varies with $x$. If you draw tangents on $f(x)$ at different points $x_1, x_2$, you get different slopes, don't you?
