Answer to Question #95908 in Differential Equations for JrS

Question

(3x+y-z)p+(x+y-z)q=2(z-y)

Answer 1

The PDE has the form

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

To solve this equation we need to find 2 independent solutions of characteristic system

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

We can use the following property:

$$\sum\limits_{k = 1}^n {\frac{{{a_k}}}{{{b_k}}}} = \frac{{\sum\limits_{k = 1}^n {{\lambda _k}{a_k}} }}{{\sum\limits_{k = 1}^n {{\lambda _k}{b_k}} }}$$

First, choose ${\lambda _1} = - 1$ , ${\lambda _2} = 3$ and ${\lambda _3} = 1$ and we get

$$- dx + 3dy + dz = 0$$

Then the first integral of the system is

$${\varphi _1} = - x + 3y + z$$

Now let's choose ${\lambda _1} = 1$ , ${\lambda _2} = - 1$ , ${\lambda _3} = 1$ and then ${\lambda _1} = 1$ , ${\lambda _2} = 1$ , ${\lambda _3} = -1$ . Because in this case $\sum\limits_{k = 1}^n {{\lambda _k}{b_k}} \ne 0$ these expressions will be equal so we get

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

$$\frac{{d(x - y + z)}}{{x - y + z}} = \frac{1}{2}\frac{{d(x + y - z)}}{{x + y - z}}$$

$$\ln \left| {x - y + z} \right| = \ln \sqrt {\left| {x + y - z} \right|}$$

Then the second integral is

$${\varphi _2} = x - y + z - \sqrt {x + y - z}$$

Now we know 2 independent integrals of the characteristic system so we can write the solution of the PDE in the form

$$F({\varphi _1};{\varphi _2}) = 0$$

or explicitly

$$F( - x + 3y + z;x - y + z - \sqrt {x + y - z} ) = 0$$

where $F$ is a smooth differentiable function.