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#### Julio

##### Member

- Feb 14, 2014

- 71

Let $z:\mathbb{R}^2\to \mathbb{R}$ an function of kind $C^2(\mathbb{R}^2)$. What transforms the equation $2\dfrac{\partial^2 z}{\partial x^2}+\dfrac{\partial^2 z}{\partial x\partial y}-\dfrac{\partial^2 z}{\partial y^2}+\dfrac{\partial z}{\partial x}+\dfrac{\partial z}{\partial y}=0$ under the change of variable $u=x+2y+2$ and $v=x-y-1$?

Hi, I have this problem I don't understand how to solve it. I have calculated the following:

$\dfrac{\partial z}{\partial x}=\dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial x}+\dfrac{\partial z}{\partial v}\dfrac{\partial v}{\partial x}=\dfrac{\partial z}{\partial u}+\dfrac{\partial z}{\partial v}.$

$\dfrac{\partial z}{\partial y}=\dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial y}+\dfrac{\partial z}{\partial v}\dfrac{\partial v}{\partial y}=2\dfrac{\partial z}{\partial u}-\dfrac{\partial z}{\partial v}.$

However, the case $\dfrac{\partial^2 z}{\partial x^2}$ I don't understand how solve, anyone can help me?

$\dfrac{\partial z}{\partial x}=\dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial x}+\dfrac{\partial z}{\partial v}\dfrac{\partial v}{\partial x}=\dfrac{\partial z}{\partial u}+\dfrac{\partial z}{\partial v}.$

$\dfrac{\partial z}{\partial y}=\dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial y}+\dfrac{\partial z}{\partial v}\dfrac{\partial v}{\partial y}=2\dfrac{\partial z}{\partial u}-\dfrac{\partial z}{\partial v}.$

However, the case $\dfrac{\partial^2 z}{\partial x^2}$ I don't understand how solve, anyone can help me?

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