Answer to Question #160927 in Differential Equations for Akash.R

Question

Solve the following equation of the form 1+p^2=qz

Answer 1

1+p²-qz=0

Therefore the Charpit's auxiliary equation is

$\frac{dp}{\frac{df}{dx}+p \frac{df}{dz}}=\frac{dq}{\frac{df}{dy}+q \frac{df}{dz}}=\frac{dz}{-p\frac{df}{dp}-q\frac{df}{dq}}=\frac{dx}{\frac{-df}{dp}}=\frac{dy}{-\frac{df}{dq}}$

or

$\frac{dp}{0-pq}=\frac{dq}{0-q^2}=\frac{dz}{-2p^2+qz}=\frac{dx}{-2p}=\frac{dy}{z}$

$\frac{dp}{-p}=\frac{dq}{-q}$

$p=q$

$\frac{2}{q^3}=y/z$

$q=\sqrt[3]{\frac{2z}{y}}$

$dz=pdx+qdy$ $dz=\sqrt[3]{\frac{2z}{y}}dx+\sqrt[3]{\frac{2z}{y}}dy$

$z=x\sqrt[3]{\frac{2z}{y}}-\frac{-1}{3}\sqrt[3]{\frac{2z}{y^4}}$