Question #93313
Find the complete integral of
p²+q²-2px-2qy+1=0
1
Expert's answer
2019-09-09T10:47:57-0400

f(x,y,z,p,q)=p2+q22px2qy+1=0f(x,y,z,p,q)=p^2+q^2-2px-2qy+1=0


dpfx+pfz=dqfy+qfz=dzpfpqfq=dxfp=dyfq\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}

dpp=dqq=dzp(px)+q(qy)=dxpx=dyqy\frac{dp}{p}=\frac{dq}{q}=\frac{dz}{p(p-x)+q(q-y)}=\frac{dx}{p-x}=\frac{dy}{q-y}

dpp=dqqlnp=lnq+lncp=cq\frac{dp}{p}=\frac{dq}{q} \Rightarrow \ln p=\ln q + \ln c \Rightarrow p=cq


f(x,y,z,p,q)=c2q2+q22cqx2qy+1=q2(c2+1)2q(cx+y)+1=0f(x,y,z,p,q)=c^2q^2+q^2-2cqx-2qy+1=q^2(c^2+1)-2q(cx+y)+1=0

q=2(cx+y)±4(cx+y)24(c2+1)2(c2+1)q=\frac{2(cx+y) \pm \sqrt{4(cx+y)^2-4(c^2+1)}}{2(c^2+1)}

cx+y=t,c2+1=acx+y=t, \,\, c^2+1=a


q=t±t2aa,p=cq=ca(t±t2a)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)=qdtdz=pdx+qdy=cqdx+qdy=q(cdx+dy)=qd(cx+y)=qdtz=t±t2aadt=t22a±1at2adtz=\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=t22a±12tt2a12aln(t2a+t)+Cz=\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=(cx+y)22a±12(cx+y)(cx+y)2a12aln((cx+y)2a+cx+y)+Cz=\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


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