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

$$solvePDE{py}^3+{qx}^2=0$$

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

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

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