Question

Solve the differential equation

(yz+z²) dx - xz dy + xy dz = 0

Answer 1

Here

$$P= yz+z^2 ,Q= -xz ,R= xy$$ $${dx \over yz+z^2}={dy \over -xz}={dz \over xy}$$

From 2nd and 3rd fraction, we get 

$${dy \over -z}={dz \over y}$$ $$d(y^2+z^2)=0$$

Integrating, we get 

$$y^2+z^2=c_1$$

If we take

$$P_1=x, Q_1=z,R_1=-z$$

then

$$PP_1+QQ_1+RR_1=xyz+xz^2-xz^2-xyz=0$$

Thus for the given system of equations, we have 

$${dx \over yz+z^2}={dy \over -xz}={dz \over xy}={xdx+zdy-zdz \over 0}$$

$${dx \over x}={dy \over z}={dz \over -z}$$

From 1st and 2nd fraction, we get 

$${dx \over x}={dz \over -z}$$ $$d(\ln|x|+\ln|z|)=0$$

Integrating, we get 

$$\ln(|x||z|)=\ln c^*$$ $$|x||z|=c_2$$

Let F be an arbitrary differentiable function, then the solution of the partial differential equation is

$$F(y^2+z^2,|x||z|)=0$$