Question

How to solve this equation with homogeneous coefficients
x(x^2+y^2)(ydx-xdy)+y^6dy=0

Answer 1

its a differential equation problem

Answer 2

i think you should be a little more specific...

Answer 3

how to solve this using the
theorem 1:if M (x,y) and N(x,y) are both homogeneous & of the same degree, the function M(x,y)/N(x,y) is homogeneous of degree zero

Answer 4

ohh that

Answer 5

& using also theorem 2 of homogeneous function
theorem 2: If f(x,y0 is homogeneous of degree zero in x and y, f(x,y) is a function of y/x alone.

Answer 6

have you tried starting this?

Answer 7

i'm confused in what to use if its y=vx or x=vy & when to substitute it..

Answer 8

i have here the steps in solving

Answer 9

you have to change it to look like Mdx + Ndy = 0 first

Answer 10

1. replace y by y=vx or x=vy
2. Factor out x or y to obtain DE in which the coefficients are functions of v alone.
3. use variable separable method. If the DE is not exact

Answer 11

change. it. to. Mdx + Ndy = 0. first.

Answer 12

how?

Answer 13

expand x(x^2 + y^2)(ydx - xdy)

Answer 14

tag me when you got the exapnsion :D

Answer 15

@lgbasallote (x^3+xy+xy^2)(xydx-x^2dy)+y^6dy=0

Answer 16

uhh
1) how did you get that
2) that's still not in the form Mdx + Ndy = 0
`

Answer 17

i distributed x.

Answer 18

@lgbasallote i really don't know how to start this can u help me? pls. can you show me me how to solve this?

Answer 19

well..this is still algebra...distribution i mean...

distribut x into (x^2 + y^2) first. that gives you (x^3 +xy^2) now FOIL it with (ydx - xdy)

you'll get x^3 ydx + xy^3 dx - x^4 dy - x^2 y^2 dy

so the whole equation would be

\[x^3 y dx + xy^3 dx - x^4 dy - x^2 y^2 dy + y^6 dy = 0\]
factor out dx and dy
\[\implies (x^3y + xy^3)dx - (x^4 + x^2y^2 + y^6)dy = 0\]
that is now in the form Mdx + Ndy = 0

Answer 20

now do your homo thingy