# Tangent to integral curve

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?

• 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).
– user65203
Jun 14, 2014 at 10:03

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).

• 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... Jun 14, 2014 at 10:24
• @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?
– mike
Jun 14, 2014 at 10:28
• yea, didn't notice that. It's ok Jun 14, 2014 at 10:29
• hmm, I actually am having a problem with the tangent. What equation do you get for it? Jun 14, 2014 at 11:50
• @MarioStoilov I added formula $y(x)$ and made sure that point $(x_0,y_0)$ is on the curve.
– mike
Jun 14, 2014 at 13:30