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Question Number 139159 by rs4089 last updated on 23/Apr/21

solve PDE         py^3 +qx^2 =0

$${solve}\:{PDE}\:\:\:\:\:\:\:\:\:{py}^{\mathrm{3}} +{qx}^{\mathrm{2}} =\mathrm{0}\:\: \\ $$

Commented by Dwaipayan Shikari last updated on 23/Apr/21

((∂^3 f(x,y))/∂y^3 )+((∂^2 f(x,y))/∂x^2 ) =0      f(x,y)=Λe^(μ(x+y))   μ^3 e^(μ(x+y)) +μ^2 e^(μ(x+y)) =0⇒μ=0  or =−1  f(x,y)=Λe^(−(x+y))

$$\frac{\partial^{\mathrm{3}} {f}\left({x},{y}\right)}{\partial{y}^{\mathrm{3}} }+\frac{\partial^{\mathrm{2}} {f}\left({x},{y}\right)}{\partial{x}^{\mathrm{2}} }\:=\mathrm{0}\:\:\:\: \\ $$$${f}\left({x},{y}\right)=\Lambda{e}^{\mu\left({x}+{y}\right)} \\ $$$$\mu^{\mathrm{3}} {e}^{\mu\left({x}+{y}\right)} +\mu^{\mathrm{2}} {e}^{\mu\left({x}+{y}\right)} =\mathrm{0}\Rightarrow\mu=\mathrm{0}\:\:{or}\:=−\mathrm{1} \\ $$$${f}\left({x},{y}\right)=\Lambda{e}^{−\left({x}+{y}\right)} \\ $$

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