Question

Find the integral surface of the PDE
(x-y)p+(y-x-z)q=z

Answer 1

ANSWER on Question #79036 – Math – Differential Equations

QUESTION

Find the integral surface of the PDE

$$(x - y)p + (y - x - z)q = z$$

SOLUTION

Consider a quasilinear equation

$$a(x, y, z)p + b(x, y, z)q = c(x, y, z).$$

By Lagrange’s method the auxiliary equations are as following:

$$\frac{dx}{a(x, y, z)} = \frac{dy}{b(x, y, z)} = \frac{dz}{c(x, y, z)}.$$

So, for the given quasilinear equation we come to the system in the symmetric form

$$\frac{dx}{x - y} = \frac{dy}{y - x - z} = \frac{dz}{z}.$$

One of a way to solve the system in symmetric form is to use the equal fractions property

$$\frac{a_1}{b_1} = \frac{a_2}{b_2} = \cdots = \frac{a_n}{b_n} = \frac{\lambda_1 a_1 + \lambda_2 a_2 + \cdots + \lambda_n a_n}{\lambda_1 b_1 + \lambda_2 b_2 + \cdots + \lambda_n b_n}$$

In our case,

Choosing $\lambda_1 = \lambda_2 = \lambda_3 = 1$ as multipliers, each fraction on (1):

$$\frac{dx}{x - y} = \frac{dy}{y - x - z} = \frac{dz}{z} = \frac{1 \cdot dx + 1 \cdot dy + 1 \cdot dz}{1 \cdot (x - y) + 1 \cdot (y - x - z) + 1 \cdot z} \rightarrow$$ $$\frac{dx}{x - y} = \frac{dy}{y - x - z} = \frac{dz}{z} = \frac{d(x + y + z)}{x - y + y - x - z + z} \rightarrow \frac{dx}{x - y} = \frac{dy}{y - x - z} = \frac{dz}{z} = \frac{d(x + y + z)}{0} \rightarrow$$ $$d(x + y + z) = 0 \rightarrow \boxed{x + y + z = C_1}$$

Take the last two fractions of (1) and using (2) we get

$$\left\{ \begin{array}{l} \frac {d y}{y - x - z} = \frac {d z}{z} \\ x + y + z = C _ {1} \end{array} \right. \to \left\{ \begin{array}{l} \frac {d y}{y - (x + z)} = \frac {d z}{z} \\ x + z = C _ {1} - y \end{array} \right. \to \frac {d y}{y - (C _ {1} - y)} = \frac {d z}{z} \to \frac {d y}{y - C _ {1} + y} = \frac {d z}{z} \to$$ $$\frac {d y}{2 y - C _ {1}} = \frac {d z}{z} \rightarrow \int \frac {d y}{y - C _ {1} + y} = \int \frac {d z}{z} \rightarrow \frac {1}{2} \cdot \ln | 2 y - C _ {1} | = \ln | z | + \ln | C _ {2} | \Bigg | \times (2) \rightarrow$$ $$\ln | 2 y - C _ {1} | = 2 \cdot \ln | z | + \underbrace {2 \cdot \ln | C _ {2} |} _ {\ln | C _ {3} |} \rightarrow \ln | 2 y - C _ {1} | - \ln | z ^ {2} | = \ln | C _ {3} | \rightarrow \ln \left| \frac {2 y - C _ {1}}{z ^ {2}} \right| = \ln | C _ {3} | \rightarrow$$ $$\frac {(2 y - C _ {1})}{z ^ {2}} = C _ {3} \rightarrow \left[\begin{array}{c}\text{Again, we use equality (2)}\\x + y + z = C _ {1}\end{array}\right]\rightarrow\frac {(2 y - (x + y + z))}{z ^ {2}} = C _ {3} \rightarrow$$ $$\frac {2 y - x - y - z}{z ^ {2}} = C _ {3} \rightarrow \boxed {\frac {y - x - z}{z ^ {2}} = C _ {3}}$$

We have found two integrals for the given equation

$$\left\{ \begin{array}{l} x + y + z = C _ {1} \\ \frac {y - x - z}{z ^ {2}} = C _ {3} \end{array} \right.$$

Therefore, any integral surface of the differential equation $(x - y)p + (y - x - z)q = z$ is described by the equation

$$\varphi \left(C _ {1}, C _ {3}\right) = 0 \rightarrow \boxed {\varphi \left(x + y + z, \frac {y - x - z}{z ^ {2}}\right) = 0}$$

where $\varphi$ is an arbitrary function, a smooth function.

ANSWER: $(x - y)p + (y - x - z)q = z \to \varphi \left(x + y + z, \frac{y - x - z}{z^2}\right) = 0.$

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