Answer to Question #93313 in Differential Equations for Subhasis Padhy

Question

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

Answer 1

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

$$\frac{dp}{f_x+pf_z}=\frac{dq}{f_y+qf_z}=\frac{dz}{-pf_p-qf_q}=\frac{dx}{-f_p}=\frac{dy}{-f_q}$$

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

$\frac{dp}{p}=\frac{dq}{q} \Rightarrow \ln p=\ln q + \ln c \Rightarrow p=cq$

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

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

$cx+y=t, \,\, c^2+1=a$

$$q=\frac{t\pm \sqrt{t^2-a}}{a}, \,\, p=cq=\frac{c}{a}(t \pm \sqrt{t^2-a})$$

$$dz=pdx+qdy=cqdx+qdy=q(cdx+dy)=qd(cx+y)=qdt$$ $$z=\int \frac{t \pm \sqrt{t^2-a}}{a} \, dt=\frac{t^2}{2a} \pm \frac{1}{a} \int \sqrt{t^2-a} \, dt$$

$$z=\frac{t^2}{2a} \pm \frac{1}{2} t \sqrt{t^2-a} \mp \frac{1}{2} a \ln (\sqrt{t^2-a}+t)+C$$

$$z=\frac{(cx+y)^2}{2a} \pm \frac{1}{2}(cx+y) \sqrt{(cx+y)^2-a} \mp \frac{1}{2} a \ln (\sqrt{(cx+y)^2-a}+cx+y)+C$$