Answer to Question #154215 in Differential Equations for K.tha

Question

bc(b-c)yzp+ca(c-a)xzq=ab(a-b)xy

Answer 1

P=bc(b-c)yz=b²cyz - bc²yz

Q=ca(c-a)xz=c²axz - ca²xz

R=ab(a-b)xy=a²bxy - ab²xy

$\frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R}$

$\frac{dx}{b^2cyz - bc^2yz}=\frac{dy}{c^2axz - ca^2xz}=\frac{dz}{a^2bxy - ab^2xy}$

Multiplyers: ax,by,cz

axdx+bydy+czdz=0

1/2(ax²+by²+cz²)=C₁

Multiplyers: -ax,-by,-cz

-axdx-bydy-czdz=0

-1/2(ax²+by²+cz²)=C₂

The general solution is,

$\phi(c_{1},c_{2})=0$

$\phi( 1/2(ax^2+by^2+cz^2),-1/2(ax^2+by^2+cz^2))=0$