Question

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

How Do I Solve This 1st Order DifEQ? Help Please!!

Answer 1

yup this 1st Order DifEQ

Answer 2

you can use y=vx (for homogeneous equations)
to solve it

Answer 3

I don't understand how you determine if its homogeneous, separable...but I understand if its an exact equation...

Answer 4

then dy =xdv +vdx

Answer 5

I have 6 problems that i have to do and i have the answers but i just dont understand how to do the work to get the answer unless its an exact equation..

Answer 6

see the sum of the powers of x and y is uniform (here its 2 ) throughout hence homogeneous

Answer 7

Great! I'm Learning Something Finally!

Answer 8

welcome

Answer 9

after using y=vx and dy =xdv +vdx
the resultant will become a seperable one

Answer 10

Is there a certain formula that is out there where i can solve these problems if they are separable or homogenous or any other thing out there?? I really want to learn this stuff

Answer 11

You cannot solve this using separation of variables.

Answer 12

Yes your right, I understand that its not an exact equation..

Answer 13

I have 6 problems that are due in 45 mins... I need help!!!

Answer 14

Let's start from the beginning

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

implies that
dy/dx = -(2x^2+xy+y^2)/(2x^2)

and hence

dy/dx = -1 - (1/2)(y/x) - (y/x)^2

Answer 15

This is what is called a homogeneous equation. It is one of the meanings of the word homogeneous for differential equations.

The substitution 'trick' for such equations is to write v = y/x and hence
y = vx
which implies
y' = v + xv'

Therefore we can rewrite the equation above as

v + xv' = - (1 + (1/2)v + v^2)

This now IS a separable equation which you can solve using standard techniques to find the function v(x). Once you have done that, substitute back v(x) = y(x)/x to solve for y(x)

Answer 16

***I dropped a half in front of the (y/x)^2 term, so

v + xv' = -(1/2) ( 2 + v + v^2 )

Answer 17

Make sense?

Answer 18

I keep getting different answers though when i do that... i have the solutions just can get there while showing my work

Answer 19

Anyone who can help solve these problems will be a life saver!

Answer 20

Well,

v + xv' = -(1/2) ( 2 + v + v^2 )

implies

-2x v' = v^2 + 3v + 2 = (v+2)(v+1)

hence separating

\[ -2\int \left( \frac{A}{v+2} + \frac{B}{v+1} \right) dv = \int \frac{dx}{x} \]

You take it from there.