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Question Number 8043 by eva suting last updated on 28/Sep/16
$${solve}\:\left({xz}+{y}^{\mathrm{2}} \right)+\left(\mathrm{yz}−\mathrm{zx}^{\mathrm{2}} \right)\mathrm{q}+\mathrm{2xy}+\mathrm{z}^{\mathrm{2}} =\mathrm{0} \\ $$
Commented by prakash jain last updated on 28/Sep/16
$$\mathrm{solve}\:\mathrm{for}\:\mathrm{which}\:\mathrm{variable}? \\ $$
Commented by eva suting last updated on 28/Sep/16
$${I}\:{dont}\:{know}\:{but}\:{i}\:{think}\:{we}\:{have}\:{to}\:{solve}\:{by}\:{charpit} \\ $$$${or}\:{lagrange}\:{auxilliary}\:{equation}\:{whichever}\:{is}\: \\ $$$${applicable} \\ $$$$ \\ $$
Commented by 123456 last updated on 29/Sep/16
$$\left({xz}+{y}^{\mathrm{2}} \right)+\left({yz}−{zx}^{\mathrm{2}} \right){q}+\mathrm{2}{xy}+{z}^{\mathrm{2}} =\mathrm{0} \\ $$$${or} \\ $$$$\left({xz}+{y}^{\mathrm{2}} \right){p}+\left({yz}−{zx}^{\mathrm{2}} \right){q}+\mathrm{2}{xy}+{z}^{\mathrm{2}} =\mathrm{0} \\ $$$$? \\ $$
Commented by 123456 last updated on 30/Sep/16
$${f}\left({x},{y},{z},{p},{q}\right)=\left({yz}−{zx}^{\mathrm{2}} \right){q}+\mathrm{2}{xy}+{z}^{\mathrm{2}} \\ $$$${f}_{{x}} =−\mathrm{2}{zxq}+\mathrm{2}{y} \\ $$$${f}_{{y}} ={zq}+\mathrm{2}{x} \\ $$$${f}_{{z}} =\left({y}−{x}^{\mathrm{2}} \right){q}+\mathrm{2}{z} \\ $$$${f}_{{p}} =\mathrm{0} \\ $$$${f}_{{q}} ={yz}−{zx}^{\mathrm{2}} \\ $$