Question

Find the complete integral of p2+q2-2px-2qy+1=0 .

Answer 1

Answer on Question #40469, Math, Differential Calculus | Equations

Find the complete integral of $p^2 + q^2 - 2px - 2qy + 1 = 0$.

**Solution.**

$$f(x, y, z, p, q) = p^2 + q^2 - 2px - 2qy + 1 = 0$$

Charpit’s auxiliary equation are:

$$\frac{dp}{f_x + p f_z} = \frac{dq}{f_y + q f_z} = \frac{dz}{-pf_p - qf_q} = \frac{dx}{-f_p} = \frac{dy}{-f_q}$$

or

$$\frac{dp}{p} = \frac{dq}{q} = \frac{dz}{p(p - x) + q(q - y)} = \frac{dx}{p - x} = \frac{dy}{q - y}$$

Taking first two fractions,

$$\frac{dp}{p} = \frac{dq}{q}$$

Integrating, $\ln p = \ln q + \ln a$ or $p = aq$.

Putting $p = aq$ into $f(x, y, z, p, q)$, we get:

$$a^2 q^2 + q^2 - 2aqx - 2qy + 1 = 0$$ $$q^2(a^2 + 1) - 2q(ax + y) + 1 = 0$$

Solve for $q$

$$q = \frac{2(ax + y) \pm \sqrt{4(ax + y)^2 - 4(a^2 + 1)}}{2(a^2 + 1)}$$

Let’s $(ax + y) = u$ and $(a^2 + 1) = b = \text{const}$

$$q = \frac{u \pm \sqrt{u^2 - b}}{b}$$ $$p = aq = \frac{a}{b} \left(u \pm \sqrt{u^2 - b}\right)$$

We know that

$$dz = pdx + qdy = aqdx + qdy = q(adx + dy) = qd(ax + y) = qdu$$

Integrating

$$z = \int \frac{u \pm \sqrt{u^2 - b}}{b} du = \frac{u^2}{2b} \pm \frac{1}{b} \int \sqrt{u^2 - b} du =$$ $$\begin{array}{l} = \frac {u ^ {2}}{2 b} \pm \frac {1}{2} u \sqrt {u ^ {2} - b} - \frac {1}{2} b \ln (\sqrt {u ^ {2} - b} + u) + C = \\ = \frac {(a x + y) ^ {2}}{2 b} \pm \frac {1}{2} (a x + y) \sqrt {(a x + y) ^ {2} - b} - \frac {1}{2} b \ln \left(\sqrt {(a x + y) ^ {2} - b} + (a x + y)\right) + C \\ \end{array}$$

Answer: $z = \frac{(ax + y)^2}{2b} \pm \frac{1}{2} (ax + y)\sqrt{(ax + y)^2 - b} - \frac{1}{2} b\ln \left(\sqrt{(ax + y)^2 - b} + (ax + y)\right) + C$