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Worked example: range of solution curve from slope field

Given the slope field of a differential equation, we can sketch various solutions to the equation. In this example, we analyze the range of a specific solution.

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  • blobby green style avatar for user earl kraft
    at Sal says it looks like its not going to actually get to 4, but how can one know that, any more than one can look at the initial numbers in a decreasing series and unknown limit and assume it will not get to zero (and therefore diverge)?
    (26 votes)
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    • boggle blue style avatar for user Bryan
      So we know that the solution is never going to cross y=4, because near y=4 the first derivative goes very close to zero, which means the function will just keep going right roughly horizontally, to meet even more horizontal slopes, etc. So there is never a negative slope near that area to push the function over the line y=4.

      How do we know that the solution never reaches 4? I'm not completely sure about this, but this is how I think of it.
      As the function gets closer and closer to y=4, it's slope goes more and more horizontal, meaning it approaches y=4 slower and slower. As the function gets infinitely closer to y=4, it's slope gets infinitely closer to zero, and so it approaches y=4 infinitely slow. So, as the domain goes to positive infinity, the function gets closer and closer to y=4 but also approaches slower and slower, and so it never really reaches y=4 (until infinity?).
      (19 votes)
  • blobby green style avatar for user oklimengyue
    OK, quite interesting, but why? And what is initial condition and what is range of solution? What are we doing here? What meaning does it have? What application could it be?
    (8 votes)
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    • primosaur ultimate style avatar for user Yeezbear
      Let me explain this to you. To start with, initial condition isn't a new concept. It is talking about a "condition" (you can think of it as a location, simply, in this particular exaple) where function is "starting" from (for example, function f(x) starts from (0,6)).

      In this chapter, We are looking at some diffrential equation (equation about functions, not variables) and are using slope field to solve it in situation like this: we are not given numerous values of the equation, (or the numerous value from f'(x)) so that we need to solve it by, basically, putting all the possible values of the function's slope in certain bounds. and using that field to find approximate value of the asked function.
      hope this helped you some bit! If not, please leave a comment below!
      (8 votes)
  • blobby green style avatar for user anel_aquino
    What if I am ask to write the equation for the integral curve that passes through that point (0,6)
    (6 votes)
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    • duskpin tree style avatar for user Brynjar Baadshaug
      The equation for this slope field is given in the previous video as
      dy/dx = y(4-y)/6
      This equation is indeed separable, but after you separate you will have to deal with a rather awkward integral if you want to find the integral curve, or the exact solution, which isn't the point of this video. Of course that doesn't stop us from solving it anyway

      ∫ 1 / (4y-y^2) dy = ∫ 1 / 6 dx ⇒ (ln(y)-ln(y-4)) / 4 = x / 6 + C

      We can plot in y(0)=6 to find C and our integral curve already, but if we want to solve for y(x) it will be a little cleaner to get rid of some fractions and logarithms first

      e^(ln(y)-ln(y-4)) = e^(4x/6 + C)
      y / (y-4) = Ce^(2x/3)

      Keep in mind that e^(a-b) ⇔ e^a / e^b. Now we can easily substitute (0,6) to find C

      6/2 = Ce^(2*0/3) = C*1
      thus C = 3

      To find y(x) we'll treat the right hand side as a function u(x) for the sake of simplicity

      y / (y-4) = u (x)
      y = u(y-4) = uy - 4u

      Move uy, extract the y and divide the rest to get the solution

      y(1-u) = - 4u
      y = 4u / (u -1)
      y(x) = C4e^(2x/3) / (Ce^(2x/3) - 1)

      Control that we have the correct answer by substituting x=0 and C=3 to get y=6
      y(0) = 12e^0 / (3e^0 -1) = 6

      Interestingly, we can see how y → 4 as x → ∞, because 4∞ / (∞ -1) = 4. And similarly, as x → -∞, y → 0. Also, notice that x ≠ ln(1/√27) as it is the vertical asymptote for this specific solution.
      (5 votes)
  • aqualine ultimate style avatar for user Faade
    When the problem gives the initial condition, can there be no points to the left of that point? For example, if we took off the domain (x >= 0) for this problem, would the range still be (4, 6] or would it be (4, ∞)? Thanks!
    (5 votes)
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  • leaf green style avatar for user Shaeekhshuvro
    At pm,Why y can't be equal to 3 or 2 or 1 I mean less then 4?
    (3 votes)
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  • blobby green style avatar for user jih373
    But how about the lines below y=4 and x>0 ? why we don't care about the lines below?
    (3 votes)
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  • female robot amelia style avatar for user Luca Valenti
    Is the fact that we must approach a slope of zero but that we can't reach it a consequence of the paradox that 3Blue1Brown describes in his Essence of Calculus playlist? The one referring to the fact that a rate of change requires two points but that a derivative is the rate of change at a single point? Does this show an inherent weakness in a calculus description of the world?
    (1 vote)
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    • leaf green style avatar for user cossine
      You seem to be referring to the numerical approximation of the derivative.

      When we are measuring something continuous there is always going be error. I wouldn't call this a weakness as this issue is always going to be there by nature You can try increasing numerical precision of variable to improve the approximation and consequently set the value of h appropriately to deal with trade off between truncation error and round off error.

      It is also worth considering using central difference as it generally performs better than forward difference.
      (2 votes)
  • duskpin seedling style avatar for user mackbe
    How do I figure out if a differential equation [ dy/dx = ((x^2 * y) + (y^2 *x)) / 3x +7 ] will have horizontal segments and when they will occur
    (1 vote)
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    • leafers ultimate style avatar for user Yoann Nouveau
      dy/dx =((x^2 * y) + (y^2 *x)) / (3x +7) will have horizontal segments when dy/dx = 0

      Which means when (yx^2+xy^2)/(3x+7)=0 we'll have horizontal segments. 1/3x+7 doesn't affect us, if 3x+7 = 0 (x=-7/3) the whole thing is undefined.
      So basically we're looking for solutions for both x and y when (yx^2+xy^2)=0
      if x=0 => y*0^2+0*y^2 = 0, that's our first solution
      if y=0 => 0*x^2 + x*0^2 = 0, our second solution
      and then (yx^2+xy^2)=0
      means that if yx^2 = -xy^2 then dy/dx =0
      divide both sides by x:
      yx = -y^2
      divide both sides by y:
      x=-y, that's our third (and last) solution!

      To summarize, dy/dx = (yx^2+xy^2)/(3x+7) will have horizontal segments if:
      x = 0, y = 0 or x = -y.

      Here's a graph showing the slope field for your equation, and the solutions for when it has horizontal segments.

      Basically the key realisation is to figure out that all you need is to find the conditions that has to be met for dy/dx to be equal to 0.

      Hope that helped.
      (2 votes)
  • male robot hal style avatar for user James
    Shouldn't have sal wrote the interval as [6,4)? why did he write it backwards? the question is asking for the value of x bigger than 0.
    (1 vote)
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  • blobby green style avatar for user Goenitz
    I am puzzled. The slope field is y(x) or not? as at (0,6) y(x) should be around -2 at it should be increasing to 0 gradient at x>=0
    (1 vote)
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Video transcript

- [Instructor] If the initial condition is zero comma six, what is the range of the solution curve y is equal to f of x for x is greater than or equal to zero? So we have a slope field here, for a differential equation. And we're saying, okay, if we have a solution where the initial condition is zero comma six, so zero comma six is part of that solution. So let's see, zero comma six, so this is part of the solution. And we want to know the range of the solution curve. So solution curve, you can eyeball a little bit by looking at the slope field. So as x, remember, x is gonna be greater than or equal to zero, so it's going to include this point right over here. And as x increases, you can tell from the slope, okay, y is gonna decrease, but it's gonna keep decreasing at a slower and slower rate. And it looks like it's asymptoting towards the line y is equal to four. So it's gonna get really, as x gets larger and larger, larger, it's gonna get infinitely close to y is equal to four, but it's not quite gonna get there. So the range, the y-values that this is going to take on, y is going to be greater than four. It's never gonna be equal to four. So I'll do, it's going to be greater than four. That's gonna be the bottom end of my range. And at the top end of my range, I will be equal to six. Six is the largest value that I am going to take on. Another way I could have written this is four is less than y is less than or equal to six. Either way, this is a way of describing the range. The y-values that the solution will take on for x being greater than or equal to zero. If they said for all x's, well, then you might have been able to go back this way and keep going. But they're saying the range of the solution curve for x is greater than or equal to zero. So we won't consider those values of x less than zero. So there you go, the curve would look something like that. And you can see, the highest value it takes on is six, and it actually does take on that value 'cause we're including x equaling zero. And then it keeps going down, approaching four, getting very, very close to four, but never quite equaling four.