Question

Find the integral curves of the following equations:dx/(x^2–y^2–yz) = dy/(x^2–y^2–zx) = dz/[z(x–y)]

Answer 1

Solution.

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

$\frac{dx-dy}{x^2-y^2-yz-(x^2-y^2-zx)}=\frac{dz}{z(x-y)}\\ \frac{d(x-y)}{x^2-y^2-yz-x^2+y^2+zx}=\frac{dz}{z(x-y)}\\ \frac{d(x-y)}{z(x-y)}=\frac{dz}{z(x-y)}\\ d(x-y)=dz\\ \int d(x-y)=\int dz\\ z=x-y+C_1.$

Then the first integral curves is

$z-x+y=C_1.$

2) Add the first expression to the second and equate to the third

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

Because

 $x-y=z-C_1.$

Then

$\frac{d(x+y)}{2(z-c_1 )(x+y)-z(x+y)}=\frac{dz}{z(z-c_1)}\\ \frac{d(x+y)}{(x+y)(2(z-c_1)-z)}=\frac{dz}{z(z-c_1)}\\ \frac{d(x+y)}{(x+y)(z-2c_1)}=\frac{dz}{z(z-c_1)}\\ \frac{d(x+y)}{(x+y)}=\frac{(z-2c_1)dz}{z(z-c_1)}\\ \frac{d(x+y)}{(x+y)}=\frac{2z-2c_1-z}{z(z-c_1)}dz\\ \frac{d(x+y)}{(x+y)}=\left(\frac{2}{z}-\frac{1}{z-c_1}\right)dz\\ \int\frac{d(x+y)}{(x+y)}=\int\left(\frac{2}{z}-\frac{1}{z-c_1}\right)dz\\ \ln|x+y|=2\ln|z|-\ln|z-c_1|+\ln|c_2|\\ x+y=\frac{z^2 c_2}{z-c_1}$

Substitute the value $c_1$

$x+y=\frac{z^2 c_2}{z-(z-x+y)}\\ x+y=\frac{z^2 c_2}{x-y}\\ x^2-y^2=z^2 c_2$

Then the second integral curves is

$\frac{x^2-y^2}{z^2}=c_2$