Question #92906
Find the complete integral of
p²+q²-2px-2qy+1 =0
1
Expert's answer
2019-08-26T10:36:46-0400

Equation is separable:

p22px=2qyq21p^2-2px=2qy-q^2-1

it follows that

p22px=a, 2qyq21=ap^2-2px=a,\ 2qy-q^2-1=a, for constant a

Solving the first equation for p

p22pxa=0p^2-2px-a=0

using quadratic formula

p12=x±x2+ap_{12}=x\pm\sqrt{x^2+a}

and the second equation

q22qy+1+a=0q^2-2qy+1+a=0

q12=y±y21aq_{12}=y\pm\sqrt{y^2-1-a}

from relation

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

dz=(x±x2+a)dx+(y±y21a)dydz=(x\pm\sqrt{x^2+a})dx+(y\pm\sqrt{y^2-1-a})dy (four possible combinations of signs + and -)

z=(x±x2+a)dx+(y±y21a)dyz=\int(x\pm\sqrt{x^2+a})dx+\int(y\pm\sqrt{y^2-1-a})dy

z=x2+y22±(x2+a)dx±(y21a)dyz=\frac{x^2+y^2}{2}\pm\int(\sqrt{x^2+a})dx\pm\int(\sqrt{y^2-1-a})dy

(x2+a)dx=1/2(x(a+x2)+alog(x+a+x2))+C1\int(\sqrt{x^2+a})dx=\\1/2 (x \sqrt{(a + x^2)} + a\cdot log(x + \sqrt{a + x^2}))+C_1

(y21a)dy=1/2(y1a+y2(1+a)log(y+1a+y2))+C2\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=x2+y22±1/2(x(a+x2)+alog(x+a+x2))±1/2(y1a+y2(1+a)log(y+1a+y2))+b,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=x2+y22±1/2(x(a+x2)+alog(x+a+x2))±1/2(y1a+y2(1+a)log(y+1a+y2))+b,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.


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