Question

Find the differential equations of the space curve in which the two families of surfaces

u = x² - y² = c₁ and v = y² - z² = c₂ intersect.

Answer 1

Ans:-

Given two families of surfaces

$u=x^2-y^2=c_1 \ \ \ \ and \ \ \ v=y^2-z^2= c_2$

For all level surface

$du=0 \to du=2xdx-2ydy=0 \ \to\\ \ \ \ \ \ \ \ \ \ \ \ \ xdx=ydy$

$\ \ \ \ dv=0 \to dv=2ydy-2zdz=0 \ \ \ \to \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ydy=zdz$

Then,

$xdx=ydy=zdz∣÷(xyz)$

$\dfrac{xdx}{xyz}=\dfrac{ydy}{xyz}=\dfrac{zdz}{xyz}$

$\dfrac{dx}{yz}=\dfrac{dy}{xz}=\dfrac{dz}{xy}$ is auxiliary equation

Conclusion

$(yz)p+(xz)q=xy$ is desired equation