Answer to Question #320276 in Differential Equations for ire

Question

find the general solutions of (x+y+2z) dz/dx+(x+y+2z)dz/dy=3z

Answer 1

$(x+y+2z)\frac{\partial z}{dx}+(x+y+2z)\frac{\partial z}{dy}=3z$

The auxiliary equations is:

$\frac{dx}{x+y+2z}=\frac{dy}{x+y+2z}=\frac{dz}{3z}$

A first characteristic equation comes from

$\frac{dx}{x+y+2z}=\frac{dy}{x+y+2z}$

$x=y+C_1$

A second characteristic equation comes from

$\frac{dx+dy-\frac43dz}{2x+2y+4z-4z}=\frac{dz}{3z}$

$\frac{d(x+y)-\frac43dz}{2(x+y)}=\frac{dz}{3z}$

Set $x+y=t$

$\frac{dt-\frac43dz}{2t}=\frac{dz}{3z}$

$3zdt-4zdz=2tdz$

$3z\frac{dt}{dz}-2t=4z$

$3\frac{dt}{dz}-\frac{2}{z}t=4$ (1)

Let’s solve the following equation

$3\frac{dt}{dz}-\frac{2}{z}t=0$

$3\frac{dt}{t}=2\frac{dz}{z}$

$3\ln t=2\ln z+3\ln{C(z)}$

$t=C(z)z^{\frac23}$

To find solution of the equation (1) we should think C is a function of z

$t'=C'z^{\frac23}+\frac23Cz^{-\frac13}$

$3(C'z^{\frac23}+\frac23Cz^{-\frac13})-2Cz^{-\frac13}=4$

$3C'z=4$

$C'=\frac43\frac1z$

$C=\frac43\ln |z|+C_2$

$t=(\frac43\ln |z|+C_2)z^{\frac23}$

$x+y=(\frac43\ln |z|+C_2)z^{\frac23}$

$C_2=(x+y)z^{-\frac23}-\frac43\ln |z|$

General solution of the PDE on the form of implicit equation:

$\Phi(C_1,C_2)=0$.