I have an equation like: $$4y'=y(x^2-4x-3)$$ and I have to find the equation to the tangent to the integral curve, which goes through a random point from the square $K =$ $\{-5≤x≤6,-6≤y≤5 \}$. I am confused by the square part. What is actually wanted here, me choosing a point that I like, or is there something more to it?

  • $\begingroup$ For a safe answer, consider that the point can be anywhere in the given square, and make sure to analyze if there are no "special behaviors" somewhere in it (like undefined tangent - as regards me, I don't see any). $\endgroup$
    – user65203
    Jun 14, 2014 at 10:03

1 Answer 1


I think that you just need to pick a random number, a, from 0 to 1 and pick another random number, b, also from 0 to 1. Then $x_0=-5+11a$ and $y_0=-6+11b$. So point $(x_0,y_0)$ is a random point in the specified square.

A line with known slope $k_0$ and going through a point $(x_0,y_0)$ is given by:

$$ y-y_0=k_0 (x-x_0) \text{ (1)}$$

The explicit expression of your curve is given by:

$$y(x)=C \exp\left(\frac{1}{12}(x^2-6x-9)\right)$$

To make sure that point $(x_0,y_0)$ is on the curve, we solve $C$ from

$$y_0=C \exp\left(\frac{1}{12}(x_0^2-6x_0-9)\right)$$ Since $$k(x)=\frac{dy(x)}{dx}=\frac{1}{4}y(x)(x^2-4x-3)$$

You may get the desired equation by setting $k_0=k(x_0)$ in (1).

  • 1
    $\begingroup$ well, apart from the $$x_0=−5+11a$$ and $$y_0=−5+11b$$ I kind of did the same thing, hope the assistant doesn't rage at me... $\endgroup$ Jun 14, 2014 at 10:24
  • $\begingroup$ @MarioStoilov Good. You are on the right track. By the way, I just corrected the value for $y_0$ according to your original question. Can you double check? $\endgroup$
    – mike
    Jun 14, 2014 at 10:28
  • $\begingroup$ yea, didn't notice that. It's ok $\endgroup$ Jun 14, 2014 at 10:29
  • $\begingroup$ hmm, I actually am having a problem with the tangent. What equation do you get for it? $\endgroup$ Jun 14, 2014 at 11:50
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    $\begingroup$ @MarioStoilov I added formula $y(x)$ and made sure that point $(x_0,y_0)$ is on the curve. $\endgroup$
    – mike
    Jun 14, 2014 at 13:30

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