Answer to Question #125700 in Differential Equations for Vicky

Question

Solve (y-z/yz)p+(z-x/xy)q=x-y/yx

Answer 1

The given problem must be, $\dfrac{y-z}{yz}p+\dfrac{z-x}{xz}q=\dfrac{x-y}{yx}$ .

This problem can be written as, $x(y-z)p+y(z-x)q=z(x-y)$.

The subsidiary equations are,

$\dfrac{dx}{x(y-z)} = \dfrac{dy}{y(z-x)} =\dfrac{dz}{z(x-y)}$

Using the multipliers 1,1,1 each of the above ratio is equal to

$\dfrac{dx+dy+dz}{x(y-z)+y(z-x)+z(x-y)}=\dfrac{dx+dy+dz}{0}$

Hence,

$\begin{aligned} dx+dy+dz&=0,\\ \text{Integrating we get,}\\ x+y+z&=c_{1} \end{aligned}$

Using multipliers $\frac{1}{x},\frac{1}{y},\frac{1}{z}$ the ratio is equal to

$\dfrac{\frac{1}{x}dx+\frac{1}{y}dy+\frac{1}{z}dz}{(y-z)+(z-x)+(x-y)}=\dfrac{\frac{1}{x}dx+\frac{1}{y}dy+\frac{1}{z}dz}{0}$

Hence,

$\begin{aligned} \frac{1}{x}dx+\frac{1}{y}dy+\frac{1}{z}dz&=0\\ \text{Integrating we get,}\\ \ln x+\ln y+\ln z&=\ln c_{2}\\ \ln (xyz)&=\ln c_{2}\\ xyz&=c_{2} \end{aligned}$

Thus, the general solution is

$\begin{aligned} \phi(c_{1},c_{2})&=0\\ \phi(x+y+z,xyz)&=0 \end{aligned}$