Question

I'm working through a mechanics problems with differential equations, and I need some help:

Answer 1

I haven't taken a differential equations class, so I'm not sure on how to approach this completely, but I think I have close to an answer. I'm not familiar with a lot of the differential equations terminology, so any help there would also be appreciated as well.

What I think I need is a function V(x,y) such that the two partials are:\[\frac{\partial V}{\partial x}=x, \frac{\partial V}{\partial y}=y \]Like I said, I haven't done a lot of differential equations, so I'm not familiar with the methods of finding these functions — the best I can do is just guess functions and none of my guesses have been fruitful so far.

Any help would be appreciated, as always.

Answer 2

Wait, I think one of my guesses works now?

\[V=\frac{1}{2}x^2+\frac{1}{2}y^2\]That was easy, and I feel dumb for not seeing it ><.

Answer 3

Actually there can be lot many answers for this .........

Answer 4

What is another possible solution?

Answer 5

V = 1/2 x^2 + 1/2 y^2 + any constant

Answer 6

It's actually OK in this case since the function V(x,y) is a potential function.

Answer 7

Any other function type, though, and I realise that there would be an infinite number of solutions.

I could easily find the constant by implementing boundary conditions, but when we're talking about potentials the potentials can be chosen arbitrarily and there's no difference in the physical solution to the problem.

Answer 8

its ok then

Answer 9

Yep! Thanks for clarifying that, though. It's not something I even thought about.

Answer 10

you are welcome : if you want to find a solution without guessing - this is how you do it...............I hope you know in partial differentiation the other variables are treated as a constant..
So dV/dx =x
=> dv = xdx
=> V = x^2/2 + f(y) + constant (Remember f(y) is like a constant)
Similarly perform with the other equation