Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ).

share|cite|improve this question
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

2 Answers 2

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

share|cite|improve this answer
Thank you very much for your help :) – Tariq Nov 5 '12 at 13:11
You are welcome :) – Mercy King Nov 5 '12 at 13:28

solve the linear equation for Y = Z² :

                   (∂Y/∂x −∂Y/∂y)/2 = (x+y)² +Y
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.