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.

  • $\begingroup$ Can you clearly recite the statement of a theorem that asserts only one solution of this equation can pass through $(x,y) = (0,0)$? Then carefully verify that the hypotheses of the recited theorem are met here... $\endgroup$ – Eric Towers Oct 4 '20 at 2:39
  • $\begingroup$ Any point on the direction field cannot have more than one slope. $\endgroup$ – Orpheus Oct 4 '20 at 2:51
  • $\begingroup$ This is not a recitation of a theorem. $\endgroup$ – Eric Towers Oct 4 '20 at 16:09

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 $x = 0$.


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?


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:

Create a new mathSE query. In that new query, provide a link to this query. Indicate that the new query is a followup to this query, re there is still an unresolved issue. Spell out the issue as clearly as you can, being sure to use very similar syntax as in your comment.

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.

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.

  • $\begingroup$ I don't understand......I equate $\frac{dy}{dx}=x^3+yx=c$ and plot the isocline and then create direction field with slope =c along the isocline and trace the integral curve such that the direction field is tangent to the integral curves. But these integral curves are not supposed to intersect or be tangential to each other because then it would violate the property that the slope at every point on the direction field will have only one slope and multiple curves intersecting or being tangential would make that property look ambiguous. $\endgroup$ – Orpheus Oct 4 '20 at 2:49
  • $\begingroup$ @Orpheus I have just added an Addendum to my answer, in response. $\endgroup$ – user2661923 Oct 4 '20 at 5:17
  • $\begingroup$ I pondered over the problem...I think the complication arises from the fact that $\frac{dy}{dx}=f'(x,y)=\frac{x^3+y}{x}$ is not continuous at x=0 but I'm not sure of it $\endgroup$ – Orpheus Oct 4 '20 at 9:06
  • $\begingroup$ @Orpheus If I were you, I would definitely create a 2nd mathSE query. $\endgroup$ – user2661923 Oct 4 '20 at 9:26
  • $\begingroup$ Thank you...I will create a new query! $\endgroup$ – Orpheus Oct 4 '20 at 9:27

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?

  • 1
    $\begingroup$ But OP is talking about different $f_1$, $f_2$ at the same points $(x,y)=(0,0)$. There are theorems saying the solution should be unique, but that's not the case here. $\endgroup$ – Teepeemm Oct 4 '20 at 1:46

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