Question

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

Answer 1

Answer:

One possibility is:

d/dx d/dy d/dx d/dy z = 0 (all those d-s are partial derivatives)

If z = yf(x) + xg(y)

then:

d/dx z = yf’(x) + g(y)

d/dy (d/dx z) = f’(x) +g’(y)

d/dx (d/dy d/dx z) = f’’(x) because g’ does not depend on x

d/dy(d/dx (d/dy (d/dx z))) = 0 because f’’ does not depend on y

Answer 2

$\frac{d^{4} z}{dy^{2} d {x}^{2} } = 0$

Step-by-step explanation:

$Given, \: \: z=yf(x)+xg(y) \\ Differentiating \: \: with \: respect \: \: to \: \: x, \\ \frac{dz}{dx} = yf'(x) + g(y) \\ again, \: \: Differentiating \: \: with \: respect \: \: to \: \: x, \\ \frac{d^{2}z }{d {x}^{2} } = yf''(x) + 0 \\ \frac{d^{2}z }{d {x}^{2} } = yf''(x) \\ Now, \: \: Differentiating \: \: with \: respect \: \: to \: \: y, \\ \frac{d^{3} z}{dyd {x}^{2} } = f'''(x) \\ again, \: \: Differentiating \: \: with \: respect \: \: to \: \: y, \\ \boxed{\frac{d^{4} z}{dy^{2} d {x}^{2} } = 0}$