Answer to Question #87047 in Differential Equations for Biraj

Question

Solve: y2(x-y)dx+x2(x-y)dy+z(x2+y2)dz=0

Answer 1

$${dx \over {y^2(x-y)}}={dy \over {x^2(x-y)}}=-{dx \over {z(x^2+y^2)}}$$

We chooze

$$P_1={1 \over z}, Q_1={1 \over z}, R_1=x-y$$

$${{y^2(x-y)} \over z}+{{x^2(x-y)} \over z}-{{(x^2+y^2)(x-y)} \over z}=0$$

Then function u₁(x, y, z) is determined as

$$u_1(x, y, z)=\int{{1 \over z}dx}+\int{{1 \over z}dy}+\int{(x-y)dx}$$

$${x \over z}+{y \over z}+(x-y)z=C_1$$

$${dx \over {y^2(x-y)}}={dy \over {x^2(x-y)}}$$

$${dx \over {y^2}}={dy \over {x^2}}$$

$$y^2dy=x^2dx$$

$$y^3=x^3+C_2$$

The equations

$${x \over z}+{y \over z}+(x-y)z=C_1, y^3=x^3+C_2$$

constitute the integral curves of the given differential equations.