Answer to Question #304320 in Differential Equations for Chromate

Question #304320

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

yf(x) + xg(y).


1
Expert's answer
2022-03-03T04:35:12-0500

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


Differentiating partially with respect to xx and yy respectively, we get


p=zx=yf(x)+g(y)(1)q=zy=f(x)+xg(y)(2)\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=2zyx=f(x)+g(y)(3)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


xp+yq=xyf(x)+xg(y)+yf(x)+xyg(y)=xy(f(x)+g(y))+yf(x)+xg(y)=xys+z(Using equation (3) and z=yf(x)+xg(y))z=xp+yqxys\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.


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