Answer to Question #252327 in Differential Equations for pde

Question

Determine the partial differential equation arising from

ax²+by²+z²=4

Answer 1

We can write the given relations as:

$$f(x, y, z)=ax^2+by^2+z^2-4=0$$

then differentiating partially with respect to $x$ and $y$ respectively, we have

$$\dfrac{\partial f}{\partial x}+\dfrac{\partial f}{\partial z}\cdot\dfrac{\partial z}{\partial x}=0 \text{ (Keeping }\dfrac{\partial y}{\partial x}=0)$$

$$\dfrac{\partial f}{\partial y}+\dfrac{\partial f}{\partial z}\cdot\dfrac{\partial z}{\partial y}=0 \text{ (Keeping }\dfrac{\partial x}{\partial y}=0)$$

Or

$$2ax+2z\cdot\dfrac{\partial z}{\partial x}=0 =>ax+zp=0$$

$$2by+2z\cdot\dfrac{\partial z}{\partial y}=0 =>by+zq=0$$

$$a=-\dfrac{zp}{x}, b=-\dfrac{zq}{y}$$

Substitute

$$-\dfrac{zp}{x}x^2-\dfrac{zq}{y}y^2+z^2-4=0$$

$$-pxz-qyz+z^2-4=0$$

which is required partial differential equation.