## chakshu 3 years ago homogeneous equations....how to find degree???

1. chakshu

can u explain by example??

2. chakshu

do we have to convert equation to y/x form everytime??

3. chakshu

yes

4. chakshu

what i know is that if we have equation like|dw:1354030150259:dw|

5. chakshu

we find degree by |dw:1354030245650:dw|

6. chakshu

its a function

7. chakshu

say f(x,y)

8. chakshu

the denominator x cancels numerator x n what we get degree as 2

9. chakshu

since x has power 2

10. chakshu

u catchin me?

11. chakshu

but in another question

12. chakshu

$x^n f1(y/x)+y^-n f2(x/y)$

13. chakshu

degree is considered both + and - n

14. chakshu

i want to know in what form of fraction do we have to exactly convert an equation to find degree?? x/y or y/x?? hope u get what m askin

15. across

A function $$f$$ is said to be homogeneous of degree $$k$$ if $$f(\alpha\mathbf{x})=\alpha^kf(\mathbf{x})$$. In your case,$f(\alpha x,\alpha y)=\frac{\alpha^3x^3+\alpha^3y^3}{\alpha x-\alpha y}=\alpha^2\frac{x^3+y^3}{x-y}=\alpha^2f(x,y).$

16. chakshu

@across thanx for ya help aneways