Question

Form partial differential equations by eliminating arbitrary functions z=yf(x)+xg(y)

Answer 1

z=yf(x)+xg(y)

Step-by-step explanation:

dz/dx=p=yf'(x)+g(y)

dz/dy =q=f(x)+xg'(y)

r=yf"(x)

s=f'(x))+g'(y)

px=xyf'(x)+xg'(y)

qy=yf(x)+xyg'(y)

px+qy=xy(f'(x)+g'(y))+xg(y)+yf(x)

px +qy=xys+z

Answer 2

We recall the concept of partial differentiation

Partial differentiation: A partial derivative of a function of several variables is its derivative with respect to one of them.

Given:

                     $z=yf(x)+xg(y)$

                    $\frac{dz}{dx}=yf'(x)+g(y)$

             $\frac{d}{dy} (\frac{dz}{dx})=f'(x)+g'(y)$

      $\frac{d}{dx} (\frac{d}{dy} (\frac{dz}{dx}))=f''(x)$

$\frac{d}{dy}( \frac{d}{dx} (\frac{d}{dy} (\frac{dz}{dx})))=0$

Thus both functions are eliminated