Question

Solve the PDE whose auxiliary equations as follows: 𝑑𝑥/ 2𝑦(𝑧 − 3) = 𝑑𝑦 /𝑦(2𝑥 − 𝑧) = 𝑑𝑧 /𝑦(2𝑥 − 3)

Answer 1

The auxiliary equations is:

$\frac{dx}{2y(z-3)}=\frac{dy}{y(2x-z)}=\frac{dz}{y(2x-3)}$

A first characteristic equation comes from

$\frac{dx}{2y(z-3)}=\frac{dz}{y(2x-3)}$

$(2x-3)dx=2(z-3)dz$

$x^2-3x+\frac94=z^2-6z+9+C_1$

$(x-\frac32)^2=(z-3)^2+C_1$

$C_1=(x-\frac32)^2-(z-3)^2$

A second characteristic equation comes from

$\frac{dz-dy}{y(2x-3-2x+z)}=\frac{dx}{2y(z-3)}$

$2d(z-y)=dx$

$2(z-y)=x+C_2$

$C_2=2z-2y-x$.

General solution of the PDE on the form of implicit equation:

$\Phi(C_1,C_2)=0$