Question

solve dx/(x^2-yz)=dy/(y^2-zx)=dz/(z^2-xy) plz answer soon

Answer 1

Given the subsidiary equations,

$\displaystyle\longrightarrow\sf{\dfrac{dx}{x^2-yz}=\dfrac{dy}{y^2-zx}=\dfrac{dz}{z^2-xy}}$

Taking the multipliers $\displaystyle\sf{y,\ z,\ x,}$ then the ratio,

$\displaystyle\longrightarrow\sf{\dfrac{y\ dx+z\ dy+x\ dz}{x^2y-y^2z+y^2z-xz^2+xz^2-x^2y}=\dfrac{y\ dx+z\ dy+x\ dz}{0}}$

Therefore,

$\displaystyle\longrightarrow\sf{y\ dx+z\ dy+x\ dz=0}$

On integration,

$\displaystyle\longrightarrow\sf{xy+yz+zx=c_1\quad\quad\dots(1)}$

Again, taking multipliers $\displaystyle\sf{xy,\ xz,\ x^2,}$

$\displaystyle\longrightarrow\sf{\dfrac{xy\ dx+xz\ dy+x^2\ dz}{x^3y-xy^2z+xy^2z-x^2z^2+x^2z^2-x^3y}=\dfrac{xy\ dx+xz\ dy+x^2\ dz}{0}}$

Therefore,

$\displaystyle\longrightarrow\sf{xy\ dx+xz\ dy+x^2\ dz=0}$

On integration,

$\displaystyle\longrightarrow\sf{\dfrac{x^2y}{2}+xyz+x^2z=c_2}$

Or,

$\displaystyle\longrightarrow\sf{x^2y+2xyz+2x^2z=2c_2=C_2\ \ (Say)\quad\quad\dots(2)}$

Finally from (1) and (2), the solution is,

$\displaystyle\longrightarrow\sf{\underline{\underline{\Phi(xy+yz+zx,\ x^2y+2xyz+2x^2z)=0}}}$

where $\displaystyle\sf{\Phi}$ is an arbitrary function.