Question

Solve the differential equation.

dx/ y²(x-y) = dy/ x²(x-y) = dz/ z(x²+y²)

Answer 1

Here

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

From 1st and 2nd fraction, we get 

$$(x^2(x-y))dx=(y^2(x-y))dy$$

Dividing by $(x-y),$ we get

$$x^2dx-y^2dy=0$$ $$d(x^3-y^3)=0$$

Integrating, we get 

$$x^3-y^3=c_1$$

If we take

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

then

$$PP_1+QQ_1+RR_1=$$ $$={1 \over y}(y^2(x-y))-{1 \over x}(x^2(x-y))+{1 \over z}(z(x^2+y^2))=$$ $$=xy-y^2-x^2+xy+x^2+y^2=0$$

Thus for the given system of equations, we have 

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

$$={dx/y-dy/x+dz/z \over 0}$$ $${dx \over y}=-{dy \over x}={dz \over z}$$

From 1st and 2nd fraction, we get

$${dx \over y}=-{dy \over x}$$ $$xdx+ydy=0$$ $$x^2+y^2=c_2$$

Let F be an arbitrary differentiable function. Then the general equation of a partial differential equation is

$$F(x^3-y^3,x^2+y^2)=0$$