Question

Find the general solution of the linear partial differential equation, yzp+zxq=xy.

(a) φ(x^2-y^2 – z^2 )=0

(b) φ(x^2-y^2, y^2-z^2 )=0

(c) φ(x^2-y^2, y^2-x^2 )=0

(d) φ(x^2-z^2, z^2-x^2 )=0

The question was posed to me in a national level competition.

This is a very interesting question from First Order Linear PDE in portion Partial Differential Equations of Engineering Mathematics

Answer 1

The correct option is (b) φ(x^2-y^2, y^2-z^2 )=0

To explain: Given: yzp+zxq=xy

Here, the subsidiary equations are, \(\frac{dx}{yz}=\frac{dy}{zx}=\frac{dz}{xy} \)

From the first two and last two terms, we get, respectively,

\(\frac{dx}{yz}=\frac{dy}{zx’}\)       or       xdx-ydy=0,          and

\(\frac{dx}{zx}=\frac{dy}{xy’}\)      or       ydx-zdy=0,

Integrating these we get two solutions

x^2-y^2=a ,                        y^2-z^2=b

Hence, the general solution of the given equation is,

φ(x^2-y^2, y^2-z^2 )=0.