Question

integral curve of dx/y^2+yz+x^2= dy/y^2-xz+x^2=dz/z(x+y)

Answer 1

Solution:-

1)Subtract first expression from the second and equate to the third;

$\frac{dx-dy}{y^2+yz+x^2-y^2+xz-x^2}$ =$\frac{dz}{z(x+y)}$

To obtain;

$\frac{dx-dy}{z(y+x)}=\frac{dz}{z(x+y)}$

Simplify to obtain;

$d(x-y)=1dz$

Integrate both sides;

$\int d(x-y)=\int 1dz$

$x-y+C=z$

$z=x-y+C_1$

Gives the first integral curve.

2)Add the third expression to the second and equate with the first one;

$\frac{dx}{y^2+yz+x^2}=\frac{dy+dz}{y^2-xz+x^2+xz+yz}$

Simplify;

$\frac{dx}{y^2+yz+x^2}=\frac{dy+dz}{y^2+yz+x^2}$

Multiple both sides with $y^2+yz+x^2$ to obtain;

$dx=dy+dz$

$dx=d(y+z)$

Integrate both sides;

$\int dx=\int d(y+z)$

$x+C_2=y+z$

$z=x-y+C_2$

Means the equation has only one integral curve;

$z-x+y=C$