Answer to Question #225421 in Differential Equations for Anuj

Question

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

Answer 1

We first subtract the first expression from the second and equate to the third expression

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

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

Simplify to get

d(x-y)=1dz

Integrate both sides

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

x-y+C=z

z=x-y+C₁ ,this is the first integral curve

Second we add the third expression to the second and then equate with the first expression

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

Simplify to get

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

Further simply by multiplying both sides with y²+yz+x² to get

dx=dy+dz=d(y+z)

Integrate both sides

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

x+C₂=y+z

z=x-y+C₂

Thus the equation only has one integral curve: z=x-y+C

Which can also be written, z-x+y=C