Question Number 139159 by rs4089 last updated on 23/Apr/21
$${solve}\:{PDE}\:\:\:\:\:\:\:\:\:{py}^{\mathrm{3}} +{qx}^{\mathrm{2}} =\mathrm{0}\:\: \\ $$
Commented by Dwaipayan Shikari last updated on 23/Apr/21
$$\frac{\partial^{\mathrm{3}} {f}\left({x},{y}\right)}{\partial{y}^{\mathrm{3}} }+\frac{\partial^{\mathrm{2}} {f}\left({x},{y}\right)}{\partial{x}^{\mathrm{2}} }\:=\mathrm{0}\:\:\:\: \\ $$$${f}\left({x},{y}\right)=\Lambda{e}^{\mu\left({x}+{y}\right)} \\ $$$$\mu^{\mathrm{3}} {e}^{\mu\left({x}+{y}\right)} +\mu^{\mathrm{2}} {e}^{\mu\left({x}+{y}\right)} =\mathrm{0}\Rightarrow\mu=\mathrm{0}\:\:{or}\:=−\mathrm{1} \\ $$$${f}\left({x},{y}\right)=\Lambda{e}^{−\left({x}+{y}\right)} \\ $$