Question #228418

Form the partial differential equation by eliminating the arbitrary function f from z=f(x2-y2)

Form the partial differential equation by eliminating the arbitrary function f from z=f(xy)

Form the partial differential equation by eliminating the arbitrary function from z2-xy=f(x/z)


1
Expert's answer
2021-08-24T02:54:46-0400

1)

Z=f(x2y2)δZδx=2xf(x2y2)δZδy=2yf(x2y2)yδZδx+xδZδy=0Z=f(x^2-y^2)\\ \dfrac{\delta{Z}}{\delta{x}}=2xf'(x^2-y^2)\\ \dfrac{\delta{Z}}{\delta{y}}=-2yf'(x^2-y^2)\\ \therefore\\ y\dfrac{\delta{Z}} {\delta{x}}+x\dfrac{\delta{Z}}{\delta{y}}=0

2)

z=f(xy)δZδx=yf(xy)δZδy=xf(xy)xδZδxyδZδy=0z=f(xy)\\ \dfrac{\delta{Z}}{\delta{x}}=yf'(xy)\\ \dfrac{\delta{Z}}{\delta{y}}=xf'(xy)\\ \therefore\\ x\dfrac{\delta{Z}}{\delta{x}}-y\dfrac{\delta{Z}}{\delta{y}}=0

3)

z2xy=f(xz)2zδzδxy=(1zxz2δzδx)f(xz)2zδzδyx=(xz2δzδy)f(xz)δzδy(xyz21)x2z2δzδx=xzz^2-xy=f(\frac{x}{z})\\ 2z\dfrac{\delta{z}}{\delta{x}}-y=(\dfrac{1}{z}-\dfrac{x}{z^2}\dfrac{\delta{z}}{\delta{x}} )f'(\frac{x}{z})\\ 2z\dfrac{\delta{z}}{\delta{y}}-x=(-\dfrac{x}{z^2}\dfrac{\delta{z}}{\delta{y}} )f'(\frac{x}{z})\\ \therefore\\ \dfrac{\delta{z}}{\delta{y}}(\dfrac{xy}{z^2}-1)-\dfrac{x^2}{z^2}\dfrac{\delta{z}}{\delta{x}}=-\dfrac{x}{z}


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