Find the general solution of the differential equation $z(px-qy)=y^2-x^2$ Find the general solution of $$z(px-qy)=y^2-x^2$$ Let $F(x,y,z,p,q)=z(px-qy)+x^2-y^2$. This gives $$F_x=zp+2x$$
$$F_y=-zq-2y$$
$$F_z=px-qy$$
$$F_p=zx$$
$$F_q=-zy$$
By Charpit's method we have $$\frac{dx}{zx}=\frac{dy}{-zy}=\frac{dz}{z(px-qy)}=\frac{dp}{-zp-2x-p^2x+pqy}=\frac{dq}{zq+2y-pxy+q^2y}$$
By equating the first two I am getting $xy=k$.
But I am not able to solve the last two. 
Thanks for the help!!
 A: Hint : $$\frac{\,d(x+y)}{z(x-y)}=\frac{\,dz}{(y^2-x^2)} \text{ , with the help of given equation. }$$
$$\implies (x+y)\,d(x+y)+z\,dz=0$$
A: First observe that
$$2z \partial_x z = \partial_x(z^2)$$
and similarly for $y$ and set
$$w := z^2$$
We get :
$$x \partial_x w - y \partial_y w = 2y^2-2x^2$$
We can define ;
$$u=ln(x)$$
$$v=ln(y)$$
and get :
$$\partial_u w - \partial_v w = 2e^{2v}-2e^{2v}$$
We can now easily see :
$$w = f(u)+g(v)$$
as a solution form and we get :
$$f'-g'= 2e^{2v}-2e^{2u}$$
So
$$f'(u)+2e^{2u} = c = g'(v)+2e^{2v}$$
for some constant $c$
And solving we get :
$$f(u)=a+cu-e^{2u}$$
$$g(v)=b+cv-e^{2v}$$
for some constants $a$ and $b$
And we finally get :
$$z^2=r+c\,ln(xy)-x^2-y^2$$
For some constant $r$
A: $z(xz_x-yz_y)=y^2-x^2$
$xz_x-yz_y=\dfrac{y^2-x^2}{z}$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=x$ , letting $x(0)=1$ , we have $x=e^t$
$\dfrac{dy}{dt}=-y$ , letting $y(0)=y_0$ , we have $y=y_0e^{-t}=\dfrac{y_0}{x}$
$\dfrac{dz}{dt}=\dfrac{y^2-x^2}{z}=\dfrac{y_0^2e^{-2t}-e^{2t}}{z}$ , we have $z^2=f(y_0)-y_0^2e^{-2t}-e^{2t}=f(xy)-y^2-x^2$ , i.e. $x^2+y^2+z^2=f(xy)$
