Abhishek Kulkarni
Posted on Tuesday, 07 January, 2003 - 03:28 pm:
Can anybody tryout this one-

Solve the following differential equation-

(x+y)dy + (x-y)dx = 0
Chenyun Yin
Posted on Tuesday, 07 January, 2003 - 06:26 pm:
some random suggestions:
suppose parametrically y=y(t), x=x(t)
then could try:
dy/dt=x-y
dx/dt=-(x+y)

then I just guessed
x=exp(-t)cos t
y=exp(-t)sin t

arbitrary eh?
Andre Rzym
Posted on Tuesday, 07 January, 2003 - 07:59 pm:
Did any of the suggestions above lead to a solution?

If not, try the following, which I think does work:

Substitute p=ln(x) ; q=y/x

I think the result is an implicit equation for x and y.

Andre
Chenyun Yin
Posted on Tuesday, 07 January, 2003 - 10:00 pm:
x=exp(-t)cos t
y=exp(-t)sin t

this is a solution isn't it? It's a spiral.
Arun Iyer
Posted on Wednesday, 08 January, 2003 - 05:33 am:
Abhishek,
simply put y=vx and everything works fine.

love arun
Andre Rzym
Posted on Wednesday, 08 January, 2003 - 08:51 am:
Chenyun,

You are absolutely correct, your solution works very nicely. You can, of course, put the constant of integration inside the sin and cos.

One could also rewrite the differential equation (DE) in polar coordinates and the equation should be easy to integrate.

Arun, your approach is the same as mine, and will lead to an implicit equation for y in terms of x. It's not obvious looking at it (well not to me anyway!) what the shape of the curve will be. In that sense, Chenyun's is much better.

There is another approach which is just based on geometry.

Define tan(j)=dy/dx, sec(q)=y/x. Then

tan(j)=(1-tan(q))/(1+tan(q))

1=(tan(q)+tan(j))/(1-tan(q)tan(j))

tan((q)+j))=1

If you think of the geometrical interpretation of this, we have an equiangular spiral. The 'tightness' of the spiral is controlled by the '1' on the RHS of the last equation. If you change that to a constant, k, and work the equations backwards, you get the DE for a spiral of arbitrary tightness.

Andre