[Differential Equations HW help] What is the "geometric interpretation of a graph of a differential Equation"?

[Getting8926]

Question 16: The geometrical interpretation of the graph of the differential equation: [math]\frac{d y}{d x}=-\frac{x}{y}[/math] is a family of

(a) Hyperbola

(b) Ellipse

(c) Parabola

(d) Circle

I don't know how to move forward, any help would be appreciated.
I am not given access to a computer during my exams, so how can I solve the 16th problem?




My Doubt: I know integral of the differential equation gives me an equation of the circle and from the Geogebra graph, I can see the graph of the differential equation. So, if a equation of a curve is differentiated, than that differential Equation's graph would represent that particular curve (parabola or ellipse you get the idea), is that how it works? As you can see in the Geogebra graph, the graph of a differential equation derived from the equation of a circle, also seems to be making a family of circles, is this how it works?

I know the differential equation defines a family of circles, but why? But what's the reasoning behind it? If a equation of a Circle is differentiated, then would the differential define a family of circles? Can you please help me understand why this is?

Geogebra Graph of the given Differential Equation y' = -x/y


And I don't know what professor means by "the geometric interpretation of a graph of a differential Equation"?

 

[Getting8926]

> I can see the graph of the differential equation.

It's not the graph of the differential Equation, it's the slope field of the differential equation and it happens to define the curve it is derived from. i.e., if you derive the DE from a circle, it would show a family of circles (as you can see in the graph). I didn't realize it then, but I realize it now.

> , the graph of a differential equation derived from the equation of a circle, also seems to be making a family of circles, is this how it works?

That's not how it works. The differential equation derived from the equation of a circle, won't give you any circles. But as mentioned before, the slope field will partially show the outlines of the circle.

> And I don't know what professor means by "the geometric interpretation of a graph of a differential Equation"?

And I still don't know what he means, I am thinking he means the geometric interpretation of the graph of the solution of the differential equation, because apart from the slope field it won't produce anything solid. So, though I still don't understand it, I derived the the solution and it described a circle and it matched the answer.

If someone has any alternative meaning of "the geometric interpretation of a graph of a differential equation", please go on and answer this questoin.

I have marked this question as solved, so I hope I won't get any eye balls.

 

[nasi112]

And I don't know what professor means by "the geometric interpretation of a graph of a differential Equation"?

The geometric interpretation of a graph of a differential Equation simply means the graph (graphs) of the solution of the differential equation.

I know the differential equation defines a family of circles, but why? But what's the reasoning behind it? If a equation of a Circle is differentiated, then would the differential define a family of circles? Can you please help me understand why this is?

Solve the differential equation and you will know why it creates a family of circles.

Imagine that you have this differential equation, \(\displaystyle \frac{dy}{dx} = -\frac{x}{y}\), and you have no idea how to graph its solution at this moment. You will solve it normally as you solve any differential equation.

\(\displaystyle \int y \ dy = \int -x \ dx\)

\(\displaystyle \frac{y^2}{2} = -\frac{x^2}{2} + d\)

Simplify it and rearrange it.

\(\displaystyle x^2 + y^2 = c\)

Note: \(\displaystyle c = 2d\) is just another constant.

Any value of \(\displaystyle c\) that will satisfy this equation is a solution (an implicit solution) to the differential equation. And it is clear that \(\displaystyle c\) cannot be a negative number. Therefore, any value of \(\displaystyle c \geq 0\) will create a circle which means the solution of the differential equaion \(\displaystyle y'(x) = -x/y\) produces a family of circles.

 

[blamocur]

I cannot tell what your professor expects you to do, but I would draw a field of tangent vectors for each point in a grid to see what the directions of the curves are ("normalized" vector field). I asked my computer to do it for two equations [imath]y^\prime = -x/y[/imath] and [imath]y^\prime = x/y[/imath] and got two graphs from which I can see that one solutions consists of circles and another of hyperbolas (see the attached). This approach can give you some ("geometric") ideas about the nature of the solutions even when the solution is not easy to find analytically.

 

[Dr.Peterson]

I searched for the phrase in Google books, and the first few hits all said things like this:

​

Perhaps your text says something similar, which supports the interpretation that they are primarily asking about the direction field. But that doesn't tell us what the teacher expects you to do without access to tools. Solving the equation is one way; quickly sketching the direction field is another. Thinking about what the equation itself means geometrically (that the slope is perpendicular to the position vector) could be another.

You may want to ask your teacher what is expected for such a question. But since it asks for the interpretation of the graph, I suspect you are expected to actually solve it.