Given, z=yf(x)+xg(y).
Differentiating partially with respect to x and y respectively, we get
pq=∂x∂z=yf′(x)+g(y)(1)=∂y∂z=f(x)+xg′(y)(2)
Differentiating equation (1) with respect to y, we get
s=∂y∂x∂2z=f′(x)+g′(y)(3)
Multiplying equation (1) by x, Multiplying equation (2) by y, and adding we get
xp+yq∴z=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))=xp+yq−xys
which is the required partial differential equation.
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