Question Number 198355 by lapache last updated on 18/Oct/23
$${solve}\:{the}\:{EDP} \\ $$$$\mathrm{1}−\:\:\:{x}\frac{\partial{f}}{\partial{x}}−{y}\frac{\partial{f}}{\partial{y}}=\mathrm{0} \\ $$$$\mathrm{2}−\:\:\:{x}\frac{\partial{f}}{\partial{x}}−\frac{\partial{f}}{\partial{y}}=\frac{{z}^{\mathrm{2}} }{{x}} \\ $$$$ \\ $$$$\mathrm{3}−\:\:{x}^{\mathrm{2}} \frac{\partial^{\mathrm{2}} {f}}{\partial{x}^{\mathrm{2}} }−{y}^{\mathrm{2}} \frac{\partial^{\mathrm{2}} {f}}{\partial{y}^{\mathrm{2}} }=\mathrm{0} \\ $$
Commented by Gracetestifier last updated on 18/Oct/23
$${please}\:{post}\:{the}\:{answer}\:{when}\:\:{it}\:{is}\:{ready} \\ $$
Commented by mokys last updated on 18/Oct/23
![2 − (dx/x) = (dy/(−1)) = ((x dz)/z^2 ) [(dx/x) = −(dy/1)] → lnx + y = a [(dx/x) = ((xdz)/z^2 )] → (dx/x^2 ) = (dz/z^2 ) → (1/z) −(1/x)= b f(a,b)=0 → f ( lnx + y , (1/z)−(1/x))=0](https://www.tinkutara.com/question/Q198364.png)
$$\mathrm{2}\:−\:\frac{{dx}}{{x}}\:=\:\frac{{dy}}{−\mathrm{1}}\:=\:\frac{{x}\:{dz}}{{z}^{\mathrm{2}} } \\ $$$$\:\: \\ $$$$\:\:\left[\frac{{dx}}{{x}}\:=\:−\frac{{dy}}{\mathrm{1}}\right]\:\rightarrow\:{lnx}\:+\:{y}\:=\:{a}\:\:\: \\ $$$$ \\ $$$$\:\:\left[\frac{{dx}}{{x}}\:=\:\frac{{xdz}}{{z}^{\mathrm{2}} }\right]\:\rightarrow\:\frac{{dx}}{{x}^{\mathrm{2}} }\:=\:\frac{{dz}}{{z}^{\mathrm{2}} }\:\rightarrow\:\frac{\mathrm{1}}{{z}}\:−\frac{\mathrm{1}}{{x}}=\:{b}\: \\ $$$$ \\ $$$$\:\:{f}\left({a},{b}\right)=\mathrm{0}\:\rightarrow\:{f}\:\left(\:{lnx}\:+\:{y}\:,\:\frac{\mathrm{1}}{{z}}−\frac{\mathrm{1}}{{x}}\right)=\mathrm{0} \\ $$$$ \\ $$