for-the-following-problems-draw-the-directional-field-associated-with-the-differential-equation-th-2

Question

For the following problems, draw the directional field associated with the differential equation, then solve the differential equation. Draw a sample solution on the directional field.
$$
y^{\prime}=\frac{1}{x}+\ln x-y, 	ext { for } x > 0
$$

Alejandro Ruiz · Accepted Answer

So they tell us that y zero equals one, which means we&#x27;re gonna be laying down here and now our slope graph looks like this. So basically means that it&#x27;s gonna look like flat here, and then it&#x27;s gonna go for it. So this is basically the shape and then you&#x27;re bounded by you&#x27;re bounded by, um, excuse or why zero and why? Because, too

Taylor Shimono · Answer

So for this problem, we&#x27;re given the initial condition. Um Y zero is equal to negative one. And so, essentially, because we&#x27;re given our direction field for this graph, all we have to do is a first find our point on the graph, um, zero negative one, which is going to be right about here. And we&#x27;re going to look at the slope of this point and we see that it&#x27;s a positive slope, a very steeply positive slope, and so we can go ahead and draw that in our graph here. And essentially, we&#x27;re going to just follow where the arrows are pointing to the right in order to get the right side of our graph. So when we do that, we see that or a graph is going to continue increasing. I&#x27;m until hits and ass in tow at Y is equal to zero. Um, and at this point, essentially, or a graph is just going to level off. So when we connect those arrows in a nice curve, it&#x27;s going to look like this. And on the left side of our graph were essentially going to follow our arrows backwards so we can see that we are going to have a really, really, really steep line. I&#x27;m at X is equal to zero. So when we follow it or graph is going to be almost vertical on the left side and our solution should look like this when we have our initial condition, why have zero is equal to your negative one?

Taylor Shimono · Answer

So for this problem to sketch the graph of our solution with the initial condition Y zero is equal to one. Essentially, what we have to do is look at the arrow that starts at the 0.1 and then we&#x27;re going to essentially follow it through the other arrows that it points towards. So when would you that we see that our initial arrow at 01 is going to be downward slope about this? And when we follow that arrow, we can see that Amar graph is going to continue decreasing. But as it approaches the X axis, it&#x27;s going to level out. Um, And as our graph goes to the left and we follow our arrows backwards, we see that our graph is going to be getting steeper. The left. So are are a graph of the solution With the initial condition. Why have zero is equal to one should look

Taylor Shimono · Answer

so because we&#x27;re given a direction, field and an initial condition for this problem, we&#x27;re essentially going to solve this problem and sketch paragraph by finding the slope at the 0.0 negative one and then following the arrows that it points to in order to trace out our graph. So if we start with the 0.0 negative one, we can see that we&#x27;re going to have a name or a positive slope that starts below the X axis, which is going to be about like this. And as we follow that arrow, we see that our graph is one to continue increasing. But it&#x27;s going to be at a smaller rate, and eventually it&#x27;s going to be almost flat along the X axis. And if we follow our arrows in the opposite direction, I&#x27;m so instead of falling where they&#x27;re pointing, we&#x27;re going to follow the opposite direction to where they&#x27;re pointing. We find that to the left or graph is going to be getting more steeply positive so further away from the X axis at a greater rate, and so are a graph of the solution. With the initial condition, wives zero is equal to negative. One should look like this

Alejandro Ruiz · Answer

Tiu this question. What do we have to do is essentially refreshed. Draw the graph. And so it&#x27;s gonna look something like this. So negative two y sets x, like with 00 Right. Then at X Y close one, it&#x27;s gonna be negative too. So something like that, they&#x27;re all the same at the same level, right? And then at three, it&#x27;s gonna be negative. 37 more. So you get the feeling right. It&#x27;s even more even more. One more as you go up. So even more, you know, kind of like that. So? So if we wanna grab it at when Weiss are Syria, Well, if why Ciric with zero, then it&#x27;s just gonna beat this red so that CNC

Problem

For the following problems, use Euler's Method wi…

Problem 279 Easy Difficulty

For the following problems, draw the directional field associated with the differential equation, then solve the differential equation. Draw a sample solution on the directional field.
$$
y^{\prime}=\frac{1}{x}+\ln x-y, \text { for } x > 0
$$

Answer

$y(x)=\ln x+C e^{-x}$

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$y(x)=\ln x+C e^{-x}$

Calculus

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Anna Marie V.

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Gilbert Strang,

Calculus