Going Back to University!

[KrazyHorse]

Originally posted by Az
IIRC, the way to go wrt

a(x,y)(du/dx)+b(x,y)(du/dy)+c(x,y)u=d(x,y)

to replace u(x,y) with w(s,t),

with:
s=x
t=ax-by

Errr...

That substitution only works when you have a and b constants, not functions of x and y

[Arrian]

I haven't gone back in several years now. Pretty much once the people I knew while I was there stopped going, I had no reason to go back anymore. I'm sure at some point a group of us will get together and go to homecoming. It's a veritable hop, skip & a jump for me. It's harder for others who are living elsewhere...

Eek, my 10-year is coming up in '08.

-Arrian

[C0ckney]

Originally posted by KrazyHorse
I can't decide if this is a bad attempt at a troll or sheer idiocy.

and this is why spink > you.

[C0ckney]

at the risk of making an on topic post...

me and boys used to come back quite a lot in the year after we left uni. spent a lot of time in our old local, the bars, the clubs, the beach. caught up with a lot of people we knew at uni and in swansea, which was nice. it's good being back with the boys, remembering the good times, getting back into the same habits, the same banter.

[Solly]

Originally posted by C0ckney
at the risk of making an on topic post...

me and boys used to come back quite a lot in the year after we left uni. spent a lot of time in our old local, the bars, the clubs, the beach. caught up with a lot of people we knew at uni and in swansea, which was nice. it's good being back with the boys, remembering the good times, getting back into the same habits, the same banter.

No doubt it'll be Wetherspoons, then Pier Pressure till 4am, then an Upper Limit brekkie to soak it all up and start again...

[Az]

Originally posted by KrazyHorse


Errr...

That substitution only works when you have a and b constants, not functions of x and y

what's the more general rule?

[molly bloom]

Originally posted by KrazyHorse
I can't decide if this is a bad attempt at a troll or sheer idiocy.

Spink, I have less respect for you than for almost anybody else here. You're in with Kidicious and Slowwhand.

Harsh words.

But why leave Ned out ?

[molly bloom]

Originally posted by Az


what's the more general rule?

Never invade Russia.

[KrazyHorse]

Originally posted by Az


what's the more general rule?

Don't know if there is a general rule for the coefficients of du/dx and du/dy dependent on a general function of both x and y...

By the way, I ****ed up. The substitution should be s = x^2/2 t = y^2/2

Sorry about that. Doing that gives you an equation of the form

du/ds + a*du/dt = f(s,t)*u + g(s,t) which is what you're looking for

Unfortunately, when I tried working that out all the way I ran into a problem. Namely, I couldn't integrate exp(sqrt(s))/sqrt(v+s) ds (with v treated as a constant)

But I tried to do it at 3:30 in the morning, so I might have ****ed up.

[KrazyHorse]

Making the substitution I described gives the following:

du/ds + du/dt = exp(sqrt(2s))/sqrt(2t) - u/sqrt(2s)

Which is a standard linear 1st order form, and has a solution by quadrature (which, as I said, was irreducible when I tried to do it)

[Lul Thyme]

Oops, thought I had it, but made a small substitution mistake., so not sure how to solve the resulting integral anymore ... let me try agian
integral of :exp(sqrt(s))/sqrt(v+s) ds
take u=sqrt(v+s)
=
integral becomes 2exp(sqrt(u^2-a)) du

Looks a bit nice but not sure how to do that.

[KrazyHorse]

Uhhhhh....I don't understand this: 2exp(u^2-a)

should be 2exp(sqrt(u^2-a)) du

Am I wrong?

EDIT: you DanSed me

Now how do you integrate what you have there?

[Lul Thyme]

I tried a few basic things and got nowhere.
I don't know much about integrating techniques though.
I don't think this yields to anything elementary (trig or by part).
I think it's probably doable analytically with something a bit more advanced. Maybe polar coordinates or what not.

[KrazyHorse]

Maybe, but I can't do it.

[Kuciwalker]

Integral solving is one of the lamer parts of mathematics. Bleh.