Answer to Question #92906 in Differential Equations for Subhasis Padhy

Question

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

Answer 1

Equation is separable:

$p^2-2px=2qy-q^2-1$

it follows that

$p^2-2px=a,\ 2qy-q^2-1=a$, for constant a

Solving the first equation for p

$p^2-2px-a=0$

using quadratic formula

$p_{12}=x\pm\sqrt{x^2+a}$

and the second equation

$q^2-2qy+1+a=0$

$q_{12}=y\pm\sqrt{y^2-1-a}$

from relation

$dz=pdx+qdy,$ where z is solution of the equation

$dz=(x\pm\sqrt{x^2+a})dx+(y\pm\sqrt{y^2-1-a})dy$ (four possible combinations of signs + and -)

$z=\int(x\pm\sqrt{x^2+a})dx+\int(y\pm\sqrt{y^2-1-a})dy$

$z=\frac{x^2+y^2}{2}\pm\int(\sqrt{x^2+a})dx\pm\int(\sqrt{y^2-1-a})dy$

$\int(\sqrt{x^2+a})dx=\\1/2 (x \sqrt{(a + x^2)} + a\cdot log(x + \sqrt{a + x^2}))+C_1$

$\int(\sqrt{y^2-1-a})dy=\\1/2(y\cdot\sqrt{-1 - a + y^2} - (1 + a) log(y + \sqrt{-1 - a + y^2}))+C_2$

$z=\frac{x^2+y^2}{2}\pm\\ 1/2 (x \sqrt{(a + x^2)} + a\cdot log(x + \sqrt{a + x^2}))\pm\\ 1/2(y\cdot\sqrt{-1 - a + y^2} - (1 + a) log(y + \sqrt{-1 - a + y^2}))+b,$

 a and b are real constants.

Answer:

$z=\frac{x^2+y^2}{2}\pm\\ 1/2 (x \sqrt{(a + x^2)} + a\cdot log(x + \sqrt{a + x^2}))\pm\\ 1/2(y\cdot\sqrt{-1 - a + y^2} - (1 + a) log(y + \sqrt{-1 - a + y^2}))+b,$

a and b are real constants.