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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 ).

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Wolframalpha can solve PDE? –  doraemonpaul Nov 5 '12 at 22:16
    
In some cases it can, see here –  Tariq Nov 8 '12 at 7:42

1 Answer 1

up vote 3 down vote accepted

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 $f(t)=\exp(A(t/2))$.

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Thank you very much for your help :) –  Tariq Nov 5 '12 at 13:11
    
You are welcome :) –  Mercy Nov 5 '12 at 13:28

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