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I need to solve this partial differential equation,

$$ Z\left(\,{\partial Z \over \partial x} - {\partial Z \over \partial y}\,\right) =\left(\, x + y\,\right)^{2} + Z^{2} $$

Wolframalpha gave the last solution, $$ Z = \pm \,\sqrt{\vphantom{\LARGE A}\,% {\rm e}^{2c_{1}\left(\,x\ +\ y\,\right)\ +\ 2x} -\left(\, x + y\,\right)^{2}\,} $$ But I'm also looking for the steps. Thanks for help.

By the way I'll start asking frequently here if you don't mind because those who ask me usually might not be good at English ( I may be a little bit better ).

Felix Marin
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Tariq
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2 Answers2

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Set $$ u=\frac{x-y}{2},\ v=\frac{x+y}{2},\ F(u,v)=Z(u+v,v-u). $$ Then \begin{eqnarray} \frac{\partial Z}{\partial x}(x,y)&=&\frac{\partial F}{\partial u}(u,v)\cdot\frac{\partial u}{\partial x}+\frac{\partial F}{\partial v}(u,v)\cdot\frac{\partial v}{\partial x}\cr &=&\frac12\left(\frac{\partial F}{\partial u}(u,v)+\frac{\partial F}{\partial v}(u,v)\right)\cr \frac{\partial Z}{\partial y}(x,y)&=&\frac{\partial F}{\partial u}(u,v)\cdot\frac{\partial u}{\partial y}+\frac{\partial F}{\partial v}(u,v)\cdot\frac{\partial v}{\partial y}\cr &=&\frac12\left(\frac{\partial F}{\partial v}(u,v)-\frac{\partial F}{\partial u}(u,v)\right). \end{eqnarray} Now the PDE reads: $$ \frac12\frac{\partial }{\partial u}F^2(u,v)=4v^2+F^2(u,v). $$ After integration we get $$ \frac12\ln(4v^2+F^2(u,v))=u+\frac12 A(v), $$ where $A$ is an arbitrary function. It follows $$ F(u,v)=\pm \sqrt{e^{2u}e^{A(v)}-4v^2}. $$ Hence $$ Z(x,y)=F(\frac{x-y}{2},\frac{x+y}{2})=\pm\sqrt{f(x+y)e^{x-y}-(x+y)^2}, $$ where $f(t)=\exp(A(t/2))$.

HorizonsMaths
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solve the linear equation for Y = Z² :

                   (∂Y/∂x −∂Y/∂y)/2 = (x+y)² +Y
Juergen Puengel
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