Question

Z^3=pqxy by jacobi's method

Answer 1

p=-u₁/u₃; q=-u₂/u₃

z³=u₁u₂xy/u₃²

z³u₃² - u₁u₂xy=0

$\frac{dx}{-u_2xy}=\frac{dy}{-u_1xy}=\frac{dz}{2z^3u_3}=\frac{du_1}{u_1u_2y}=\frac{du_2}{u_1u_2x}=\frac{du_3}{-3z^2u_3^2}$

u₁dx=u₂dy=a

$\frac{dz}{2z}= \frac{du_3}{-3u_3}$

$\frac{log(z)}{2}= \frac{log(u_3)}{-3}$

u₃=ze^-3/2+b

u=2a+(ze^-3/2 + b)dz=2a+z²e^-3/2/2+bz+c

u=c

2a+z²e^-3/2/2+bz=0

$z=\frac{-b^+_- \sqrt{b^2-4ae^{-3/2}}}{e^{-3/2}}$