Partial differential equation in Lagrange form.

Question

Solve the pde.

$$(my(x+y)-nz^2)p-(lx(x+y)-nz^2)q=(lx-my)z$$

Using method of multipliers$(1,1,(x+y)/z)$ I got one of the solutions $$(x+y)z=c_1$$

I can't figure out what the other solution might be.

PS: $p$ and $q$ are pde wrt to x and y.

Answer 1

$$(my(x+y)-nz^2)p-(lx(x+y)-nz^2)q=(lx-my)z$$ $$\frac{dx}{my(x+y)-nz^2}=\frac{dy}{-lx(x+y)+nz^2}=\frac{dz}{(lx-my)z}=$$ $$=\frac{dx+dy+\frac{x+y}{z}}{my(x+y)-nz^2-lx(x+y)+nz^2+\frac{x+y}{z}(lx-my)z}=\frac{dx+dy+\frac{x+y}{z}dz}{0}$$ $$\frac{dx+dy}{x+y}+\frac{dz}{z}=0$$ $$\ln|x+y|+\ln|z|=\text{constant}$$ $$(x+y)z=c_1$$ I agree with your first characteristic equation.

Second characteristic equation, with multipliers $(lx,my,0)$ : $$\frac{lxdx+mydy}{lxmy(x+y)-lxnz^2-mylx(x+y)+mynz^2}=\frac{lxdx+mydy}{-(lx-my)z^2} =\frac{dz}{(lx-my)z}$$ $$lxdx+mydy+zdz=0$$ $$lx^2+my^2+z^2=c_2$$