Answer to Question #135620 in Differential Equations for khumdile

Question

find the partial differential equations arising from z=xy+f(x^2+y^2)

Answer 1

$\displaystyle z = xy + f(x^2 + y^2)\\ \textsf{Consider} \hspace{0.1cm} u = f(x^2 + y^2)\\ u = f(v), \hspace{0.1cm} v = x^2 + y^2\\ \frac{\partial u}{\partial v} = f'(v)\\ \therefore \frac{\partial u}{\partial x} = \frac{\partial u}{\partial v} \times \frac{\partial v}{\partial x}\hspace{0.1cm} \& \hspace{0.1cm} \frac{\partial u}{\partial y} = \frac{\partial u}{\partial v} \times \frac{\partial v}{\partial y}\\ \frac{\partial u}{\partial v} = f'(x^2 + y^2) \\ \frac{\partial v}{\partial x} = 2x \hspace{0.1cm} \& \hspace{0.1cm} \frac{\partial v}{\partial y} = 2y\\ \Rightarrow \frac{\partial z}{\partial x} = y + 2xf'(x^2 + y^2) \hspace{0.1cm}\&\hspace{0.1cm}\\ \frac{\partial z}{\partial y} = x + 2yf'(x^2 + y^2)\\ y\frac{\partial z}{\partial x} = y^2 + 2xyf'(x^2 + y^2) \hspace{0.5cm} (\textit{1}) \\ x\frac{\partial z}{\partial y} = x^2 + 2xyf'(x^2 + y^2)\hspace{0.5cm}(\textit{2})\\ \textsf{Subtracting}\hspace{0.1cm} (\textit{2}) \hspace{0.1cm} \textsf{from}\hspace{0.1cm} (\textit{1}), \hspace{0.1cm} \textsf{we have;}\\ y\frac{\partial z}{\partial x} - x\frac{\partial z}{\partial y} = y^2 - x^2\\ \therefore y\frac{\partial z}{\partial x} - x\frac{\partial z}{\partial y} = y^2 - x^2\\ \textsf{is a partial differential}\\\textsf{equation arising from}\hspace{0.1cm}z$