Homogeneous equation

[amb123]

(x^2 + y^2)dx + (2xy)dy = 0

I get y = sqrt((kx^5 + x^2)/3) Where k = c2 cubed, and c2 = ln(c) so k = 3ln(c)

But, the answer the teacher gave is (x^2)(y^3) - x - ln(y) = c I can't come up with anything remotely close. I know this isn't in a pretty LaTeX form, but I am new and haven't figured it out yet. Also, i'm not sure how to get my computer to read LaTeX, so if there is a program I need to d/l, can someone link me?

Any help would be great !
A

[Damned charming :)]

just see the sticky thread on using tex basically you just put tex *forward slash* tex in square brackets around
around equations and use *backward slash* sqrt instead of sqrt

QUOTE -
x [\tex]
[tex] (x^2 + y^2)dx + (2xy)dy = 0


I get y = \sqrt((kx^5 + x^2)/3) Where k = c2 cubed, and c2 = ln(c) so k = 3ln(c)

But, the answer the teacher gave is (x^2)(y^3) - x - ln(y) = c I can't come up with anything remotely close.



to answer your question I think it uses the method of exact differential equations just look it up on the web SOS would be a good start. However when I Use exact differential equations I get a totally different answer but it works when I sub things back in.

[Damned charming :)]

Thats funny I have not had a problem with latex before. I will insert spaces to show the simple things I sometimes type and it still does not work

[ tex ] x [ /tex ]
x

[JonF]

i get a different answer from both of you

(x^2 + y^2)dx + (2xy)dy = 0
Is an exact homogenous d.e.

So you take the anti-partial derivative in respect to y of your dy term:
xy^2 + h(x).
this is your solution, but you need to figure out what that left over function of x is

to do that you first take the partial of that in respect to x
y^2 + h’(x).

you know this function must be equal to your dx term since the d.e. is exact. So…

y^2 + h’(x) = x^2 + y^2
h’(x) = x^2
anti-differentiate
h(x) = (x^3)/3 + C

so your solution is
xy^2 + (x^3)/3 = C

[JonF]

you have your slashes going the wrong dirrection in the [ \tex ]

[Damned charming :)]

That was the answer I got but did not mention. It is right because you can implicitly differentiate and get back to the original differential equation.

I thought you used backslashes in latex but to go into tex mode the standard VB code is to use a forward slash the same way as bold font colour ect.

*tries [ tex ] x [ \tex]*
x [\tex]
tries [ tex ] x [ /tex ]*
[tex] x

[JonF]

The spaces were there so it didn't think I was doing latex and not show you what I was trying to convey. Proper usage is (tex) (\tex) with ] [ instead of ) (

[Muzza]

Proper usage is (tex) (\tex) with ] [ instead of ) (


Funny how this (http://www.physicsforums.com/showthread.php?t=8997) thread says otherwise ;)

[JonF]

sorry, guess i got mixed up. But hey atleast i solved the d.e. right... :bugeye: