OK, I got an exam in about a week, and there is a point that I don't really got my head around yet. Our professor likes to give for example three pictures and one differential equation.

The question now is which of these pictures approximates the given equation.

How would I go and start such a task? These are very simple equations and most will probably find this trivial, but as someone in second semester I always get stuck on these kind of questions.

Examples would be:

$$ y^\prime = y + 1 $$

$$ y^\prime = -2xy^2 $$


Enter image description here

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    $\begingroup$ These tasks are meant to be solved within a minute, so most of these should be solvable with a sharp look as our Professor likes to say. Calculating them completely takes way to much time, that i dont have. $\endgroup$
    – joachim
    Sep 17, 2016 at 13:20
  • $\begingroup$ That was mentioned yes. $\endgroup$
    – joachim
    Sep 17, 2016 at 13:43
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    $\begingroup$ Why do you need to visualize differential equations? $\endgroup$
    – Asaf Karagila
    Sep 18, 2016 at 5:04
  • $\begingroup$ I dont really know why, but its not a bad skill to have so you have a rough approximation in your head when you see something, but i just started studying last Semester so we will see i guess. $\endgroup$
    – joachim
    Oct 5, 2016 at 8:16
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    $\begingroup$ @joachim what is the book that contains the pictures in the post? $\endgroup$
    – Black Mild
    Sep 24, 2021 at 10:55

2 Answers 2


For $y^{\prime} =y+1$, the slope of a solution must be negative at any point where $y<-1$ and positive otherwise. So in your pictures, this eliminates the straight line and the parabola.

For $y^{\prime} = -2xy^2$, note that $y^2$ is non-negative, so $-2xy^2$ is negative in if $x>0$ and positive if $x<0$. Only one of the pictures shows a solution which is increasing for $x<0$ and decreasing for $x>0$.

For these sorts of problems, there's probably no single technique, but in general, think about what slope of the solution is in each quadrant. If $y^{\prime} = f(x,y)$, it might be helpful to sketch the curve $f(x,y) =0$ and then think about what happens on each side of that curve. It could help you eliminate a picture or two.

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    $\begingroup$ So to be clear, i could just try out some values and see what happens, like i would with every normal equation? $\endgroup$
    – joachim
    Sep 17, 2016 at 13:47
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    $\begingroup$ Yes. This is the point about the "slope field" or "direction field" that folks have mentioned. You plug in several values for $(x,y)$ and see what it says about $y^{\prime}$. But I think the intention of the problems are that you do this in your head. $\endgroup$
    – B. Goddard
    Sep 17, 2016 at 13:53
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    $\begingroup$ @JoachimGotzes Well, it is a partial yes: you could insert some values for $x$, but then you do not know the corresponding $y$ until you solve the ODE (including initial value), so you cannot calculate what value is resulting for $y'$. The only way is the one suggested by B. Goddard: essentially, proceed by "quadrants", what sign has $y'$ if $x<0$ and $y<0$, etc., or fixing some values for $x$ and sign for $y$ .. $\endgroup$
    – G Cab
    Sep 17, 2016 at 14:04
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    $\begingroup$ @JoachimGotzes - The trick is to pick good values to try. You can gain experience by doing practice problems, but a systematic approach would be to try quadrants (as discussed) or look for key values of x and y (e.g. x = 0, y = 0, other values where terms cancel). $\endgroup$
    – Hao Ye
    Sep 18, 2016 at 8:13

A slope field for the differential equation $$ \frac{dy}{dx} = F(x, y) \tag{1} $$ is a diagram having line segments of slope $F(x, y)$ placed at $(x, y)$, usually for points lying in a rectangular grid. Geometrically, a solution of (1) is a graph $y = f(x)$ "tangent to the field", i.e., satisfying $$ f'(x) = F\bigl(x, f(x)\bigr). $$

With a bit of practice, a function $F(x, y)$ and a plot of a slope field can be matched fairly easily; just match values of $F$ (say, points where $F = 0$) with segments having that slope.

Slope field of $y' = -2xy^{2}$ Slope field of $y' = y\sin(2\pi x)$ Slope field of $y' = 1 - y^{2}$

  • $\begingroup$ That's true we did that at some point, but putting a slope field for every equation takes a lot of time, the ~ time to solve this for 3-4 equations is setup point wise with 3 minutes. So 1 minute per equation. $\endgroup$
    – joachim
    Sep 17, 2016 at 13:51
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    $\begingroup$ @JoachimGotzes But you are given the graphs, so checking for $(x,y)=(0,y)$ for example, can be easily done. $\endgroup$ Sep 17, 2016 at 14:39

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