Question

pz+qy=pq. solve by charpits method​

Answer 1

Answer:

By Charpit's Method, the problem has been solved. The solution includes auxiliary equation with solution.

Answer 2

Answer:

The given equation is proved by the charpits method

Step-by-step explanation:

$\mathrm{px}+\mathrm{qy}-\mathrm{pq}=0-----(1)$

$\Rightarrow \mathrm{f}=\mathrm{px}+\mathrm{qy}-\mathrm{pq}$

Auxiliary equation:

$&\frac{\mathrm{dp}}{\mathrm{f}_{\mathrm{x}}+\mathrm{pf}_{\mathrm{z}}}=\frac{\mathrm{dq}}{\mathrm{f}_{\mathrm{y}}+\mathrm{q \textrm {f } _ { z }}}=\frac{\mathrm{dz}}{-\mathrm{f}_{\mathrm{p}}-\mathrm{q \textrm {f } _ { \mathrm { q } }}}=\frac{\mathrm{dx}}{-\mathrm{f}_{\mathrm{p}}}=\frac{\mathrm{dy}}{-\mathrm{f}_{\mathrm{q}}}$

$&\Rightarrow \frac{\mathrm{dp}}{\mathrm{p}}=\frac{\mathrm{dq}}{\mathrm{q}}=\frac{\mathrm{dz}}{-\mathrm{p}(\mathrm{x}-\mathrm{q})-\mathrm{q}(\mathrm{y}-\mathrm{p})}=\frac{\mathrm{dx}}{-(\mathrm{x}-\mathrm{q})}=\frac{\mathrm{dy}}{-(\mathrm{y}-\mathrm{p})}$

$&\Rightarrow \frac{\mathrm{dp}}{\mathrm{p}}=\frac{\mathrm{dq}}{\mathrm{q}} \& \mathrm{dz}=\mathrm{pdx}+\mathrm{qdy}-----[\mathrm{A}]$

$&\square \frac{\mathrm{dp}}{\mathrm{p}}=\frac{\mathrm{dq}}{\mathrm{q}} \Rightarrow \int \frac{1}{\mathrm{p}} \mathrm{dp}=\int \frac{1}{\mathrm{q}} \mathrm{dq} \Rightarrow \ln \mathrm{p}=\ln \mathrm{q}+\ln \mathrm{a} \\&\Rightarrow \mathrm{p}=\mathrm{qa}-----[2]$

$\rightarrow$ By [1] and [2], We have

$\begin{aligned}&\mathrm{px}+\mathrm{qy}-\mathrm{pq}=0 \Rightarrow \mathrm{qax}+\mathrm{qy}-\mathrm{qaq}=0 \Rightarrow \mathrm{ax}+\mathrm{y}-\mathrm{aq}=0 \\&\Rightarrow \mathrm{q}=\frac{\mathrm{ax}+\mathrm{y}}{\mathrm{a}}\end{aligned}$

$\rightarrow By [2]$, We have

$\Rightarrow \mathrm{p}=\mathrm{ax}+\mathrm{y}$

$\rightarrow By 2^{\text {nd }}$ equation of $[A]$, We have

$\begin{aligned}&\Rightarrow \mathrm{dz}=(\mathrm{ax}+\mathrm{y}) \mathrm{dx}+\frac{\mathrm{ax}+\mathrm{y}}{\mathrm{a}} \mathrm{dy} \\&\Rightarrow \frac{1}{\mathrm{ax}+\mathrm{y}} \mathrm{dz}=\mathrm{dx}+\frac{1}{\mathrm{a}} \mathrm{dy} \\&\Rightarrow \int \frac{1}{\mathrm{ax}+\mathrm{y}} \mathrm{dz}=\int \mathrm{dx}+\int \frac{1}{\mathrm{a}} \mathrm{dy} \\&\Rightarrow \ln (\mathrm{ax}+\mathrm{y})=\mathrm{x}+\frac{\mathrm{y}}{\mathrm{a}}+\mathrm{b}\end{aligned}$