Equation is separable:
p2−2px=2qy−q2−1
it follows that
p2−2px=a, 2qy−q2−1=a, for constant a
Solving the first equation for p
p2−2px−a=0
using quadratic formula
p12=x±x2+a
and the second equation
q2−2qy+1+a=0
q12=y±y2−1−a
from relation
dz=pdx+qdy, where z is solution of the equation
dz=(x±x2+a)dx+(y±y2−1−a)dy (four possible combinations of signs + and -)
z=∫(x±x2+a)dx+∫(y±y2−1−a)dy
z=2x2+y2±∫(x2+a)dx±∫(y2−1−a)dy
∫(x2+a)dx=1/2(x(a+x2)+a⋅log(x+a+x2))+C1
∫(y2−1−a)dy=1/2(y⋅−1−a+y2−(1+a)log(y+−1−a+y2))+C2
z=2x2+y2±1/2(x(a+x2)+a⋅log(x+a+x2))±1/2(y⋅−1−a+y2−(1+a)log(y+−1−a+y2))+b,
a and b are real constants.
Answer:
z=2x2+y2±1/2(x(a+x2)+a⋅log(x+a+x2))±1/2(y⋅−1−a+y2−(1+a)log(y+−1−a+y2))+b,
a and b are real constants.
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