Differential equation of vector field

[so_gr_lo]

Homework Statement

Write the vector field F = [-4, -xcos(5x)] as a differential equation (dy/dx)

Relevant Equations

Dy/dx = dF/dx x dy/dF

I was thinking of using the chain rule with

dF/dx = 0i + (3xsin(3x) - cos(3x))j

and

dF/dy = 0i + 0j

but dF/dy is still a vector so how can it be inverted to get dy/dF ?

what are the other methods to calculate this?

 

[Orodruin]

A vector field is not a differential equation.

What you can do is write a differential equation that describes the flow lines of the vector field. Is this your purpose?

 

[so_gr_lo]

Yeah I need a differential equation for the field lines

 

[Orodruin]

So, by definition the field lines of $\vec F$ satisfy $d\vec x/dt = \vec F$ or, in components,
$$
\frac{dx}{dt} = F_1, \quad \frac{dy}{dt} = F_2,
$$
where $F_i$ are the components of $\vec F$. What does this tell you about the field lines in terms of $dy/dx$?

 

[so_gr_lo]

dy/dx = dy/dt x dt/dx = F₂/F₁ ?

 

[so_gr_lo]

Or that F = dx/dt + dy/dt

 

[Orodruin]

One advice that I generally give my students when they write two different answers is to think twice about what they write down. It often helps clearing your thoughts and thinking about the reasoning behind what each statement would be. So what would be your reasoning behind these statements?

 

[so_gr_lo]

I gave 2 different statements because I’m not sure if I am supposed to use the chain rule or not. The problem is that I don’t know how to turn the vector into a scalar so that I can write it as a differential.

 

[Orodruin]

I gave 2 different statements because I’m not sure if I am supposed to use the chain rule or not. The problem is that I don’t know how to turn the vector into a scalar so that I can write it as a differential.

You should think less of what you are ”supposed” to do and more in terms of what tools you can use to achieve your goal. What does the chain rule tell you? Is that useful in this case?

 

[so_gr_lo]

The chain rule allows you to deal with composite functions, but since I don’t actually have the y and x components written in t explicitly maybe it’s not necessary. I think the flow lines need to equal F at each point since they are the tangent vector. In that case dy/dx = y’(t)/x’(t) then dy/dx = 1/2 xcos(5x). Perhaps this is the correct appproach? In this case I just set x = -4i and y = -xcos(5x)j and ignore the unit vectors.