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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB>Unit 7

Lesson 4: Reasoning using slope fields

# 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.

## Want to join the conversation?

• 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)? • 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?).
• What if I am ask to write the equation for the integral curve that passes through that point (0,6) • 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.
• 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? • 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.
• 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! • At pm,Why y can't be equal to 3 or 2 or 1 I mean less then 4? • But how about the lines below y=4 and x>0 ? why we don't care about the lines below? • 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) • 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
simplifying:
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.
https://www.desmos.com/calculator/ql5onaikzl

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.   