Question

Find the equation of the integral surface of the differential equation (x^2-yz)p+(y^2-zx)q= z^2-xy which passes through the line x =1, y =0.

Answer 1

Answer on Question #66339 – Math – Differential Equations

Question

Find the equation of the integral surface of the differential equation

$$(x^2 - yz)p + (y^2 - zx)q = z^2 - xy$$

which passes through the line $x = 1, y = 0$.

Solution

The equation of characteristics:

$$\frac{dx}{x^2 - yz} = \frac{dy}{y^2 - zx} = \frac{dz}{z^2 - xy}$$

Then

$$\dot{x} - \dot{y} = x^2 - y^2 + z(x - y) = (x - y)(x + y + z);$$ $$\dot{z} - \dot{x} = z^2 - x^2 + y(z - x) = (z - x)(x + y + z);$$ $$\dot{z} - \dot{y} = z^2 - y^2 + x(z - y) = (z - y)(x + y + z);$$ $$\frac{d(x - y)}{d(z - x)} = \frac{x - y}{z - x};$$ $$\frac{d(x - y)}{x - y} = \frac{d(z - x)}{z - x};$$ $$\ln |x - y| = \ln |z - x| + \ln c_1;$$ $$c_1 = \frac{x - y}{z - x}.$$

In the same manner

$$\frac{d(x - y)}{d(z - y)} = \frac{x - y}{z - y};$$ $$\frac {d (x - y)}{x - y} = \frac {d (z - y)}{z - y};$$ $$\ln | x - y | = \ln | z - y | + \ln c _ {2};$$ $$c _ {2} = \frac {x - y}{z - y};$$ $$\frac {d (z - x)}{d (z - y)} = \frac {z - x}{z - y};$$ $$\frac {d (z - x)}{z - x} = \frac {d (z - y)}{z - y};$$ $$\ln | z - x | = \ln | z - y | + \ln c _ {3};$$ $$c _ {3} = \frac {z - x}{z - y}.$$

For $x = 1, y = 0$ :

$$c _ {1} = \frac {1}{z - 1} ; \quad c _ {2} = \frac {1}{z} ; \quad c _ {3} = \frac {z - 1}{z}.$$

There are infinitely many solutions

$$z = c; \quad c \text{ is an arbitrary real constant}.$$

Answer: $z = c$; $c$ is an arbitrary real constant.

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