Question

Find the integral surface of the linear partial differential equation
x(y2+z)p-y(x2+z)q=(x2-y2)z,
which contains the straight line x+y=0 , z=1 .

Answer 1

Answer on Question #40467, Differential Calculus, Equations

Find the integral surface of the linear partial differential equation

$$x (y ^ {2} + z) p - y (x ^ {2} + z) q = (x ^ {2} - y ^ {2}) z,$$

which contains the straight line $x + y = 0$, $z = 1$.

Solution.

Given

$$x (y ^ {2} + z) p - y (x ^ {2} + z) q = (x ^ {2} - y ^ {2}) z$$

Lagrange's auxiliary equations of (1) are

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

thus, the solution is the system of equations:

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

Taking $t$ as a parameter, the given equation of the straight line $x + y = 0$, $z = 1$ can be put in parametric form $x = t$, $y = -t$, $z = 1$.

Using this,

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

may be re-written as

$$\left\{ \begin{array}{c} - t ^ {2} = c _ {1} \\ 2 t ^ {2} - 2 = c _ {2} \end{array} \right.$$

Eliminating $t$ from the equations, we have

$$\left\{ \begin{array}{l} 2 (- c _ {1}) - 2 = c _ {2} \\ 2 c _ {1} + c _ {2} + 2 = 0 \end{array} \right.$$

Putting values of $c_{1}$ and $c_{2}$, the desired integral surface is

$$2 x y z + x ^ {2} + y ^ {2} - 2 z + 2 = 0$$

Answer: $2xyz + x^{2} + y^{2} - 2z + 2 = 0$