Answer to Question #295785 in Differential Equations for Brian

Question

(y+z)p+(z+x)q=x+y

Answer 1

$\displaystyle (y+z)p+(z+x)q=x+y$

The auxiliary equations are;

$\displaystyle \frac{dx}{y+z}=\frac{dy}{z+x}=\frac{dz}{x+y}$

From the first two equation, we have;

$\displaystyle \frac{dx}{y+z}=\frac{dy}{z+x}\\$

integrating both sides yields;

$\displaystyle \Rightarrow\frac{x}{y+z}=\frac{y}{z+x}+a$, where a is an arbitrary constant

$\displaystyle \Rightarrow\frac{x}{y+z}-\frac{y}{z+x}=a\cdots\cdots\cdots\cdots(1)$

From the last two equation, we have;

$\displaystyle \frac{dy}{z+x}=\frac{dz}{x+y}$

integrating both sides;

$\displaystyle \frac{y}{z+x}=\frac{z}{x+y}+b$, where a is an arbitrary constant

$\displaystyle \Rightarrow\frac{y}{z+x}-\frac{z}{x+y}=b\cdots\cdots\cdots\cdots(2)$

from (1) and (2), we have the solution as;

$\displaystyle f\left(\frac{x}{y+z}-\frac{y}{z+x},\ \frac{y}{z+x}-\frac{z}{x+y}\right)=0$