Question

Form a partial differential equation by eliminating the functions f and g from z =

yf(x) + xg(y).

Answer 1

Given, $z = yf(x) + xg(y).$

Differentiating partially with respect to $x$ and $y$ respectively, we get

$\begin{aligned} p &= \dfrac{\partial z}{\partial x} = yf'(x)+g(y) \qquad (1)\\ q &= \dfrac{\partial z}{\partial y} = f(x)+xg'(y) \qquad (2)\\ \end{aligned}$

Differentiating equation (1) with respect to y, we get

$s= \dfrac{\partial^{2} z}{\partial y \partial x} = f'(x) + g'(y) \qquad(3)$

Multiplying equation (1) by x, Multiplying equation (2) by y, and adding we get

$\begin{aligned} xp +yq &= xyf'(x)+xg(y) + yf(x)+xyg'(y) \\ &= xy(f'(x)+g'(y)) + yf(x)+xg(y)\\ &= xys+z \qquad(\text{Using equation (3) and}~ z = yf(x) + xg(y))\\ \therefore z &=xp + yq - xys \end{aligned}$

which is the required partial differential equation.