Answer to Question #142774 in Differential Equations for Srilatha Inakollu

Question

Pde(b-c/a)yzp+(c-a/b)zxq=(a-b/c)xy

Answer 1

$$Solution$$

$\frac {dx}{(b-c/a)yz}=\frac {dy}{(c-a/b)xz}=\frac {dz}{(a-b/c)xy} \implies$$\frac {xdx+ydy+zdz}{xyz(b-c/a+c-a/b+a-b/c)}=\frac {dx}{(b-c/a)yz}$

$\frac {x^2+y^2+z^2}{b-c/a+c-a/b+a-b/c}=\frac {x^2}{b-c/a}+C_1$

$\frac {xdx+ydy+zdz}{xyz(b-c/a+c-a/b+a-b/c)}=\frac {dy}{(c-a/b)xz}$

$\frac {x^2+y^2+z^2}{b-c/a+c-a/b+a-b/c}=\frac {y^2}{c-a/b}+C_2$

$\frac {xdx+ydy+zdz}{xyz(b-c/a+c-a/b+a-b/c)}=\frac {dz}{(a-b/c)xy}$

$\frac {x^2+y^2+z^2}{b-c/a+c-a/b+a-b/c}=\frac {z^2}{a-b/c}+C_3$