Question #289455

Using Charpit’s method, solve:


P²+ q² -2px -2qy +1=0

1
Expert's answer
2022-01-28T12:50:37-0500

Solution:

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

Charpit's auxiliary equation are:

dpfx+pfz=dqfy+qfz=dzpfpqfq=dxfp=dyfq\frac{d p}{f_{x}+p f_{z}}=\frac{d q}{f_{y}+q f_{z}}=\frac{d z}{-p f_{p}-q f_{q}}=\frac{d x}{-f_{p}}=\frac{d y}{-f_{q}}

or

dpp=dqq=dzp(px)+q(qy)=dxpx=dyqy\frac{d p}{p}=\frac{d q}{q}=\frac{d z}{p(p-x)+q(q-y)}=\frac{d x}{p-x}=\frac{d y}{q-y}

Taking first two fractions,

dpp=dqq\frac{d p}{p}=\frac{d q}{q}

Integrating, lnp=lnq+lnaorp=aq.\ln p=\ln q+\ln a \quad or \quad p=a q.

Putting p=aqp=a q into f(x, y, z, p, q), we get:

a2q2+q22aqx2qy+1=0q2(a2+1)2q(ax+y)+1=0\begin{aligned} &a^{2} q^{2}+q^{2}-2 a q x-2 q y+1=0 \\ &q^{2}\left(a^{2}+1\right)-2 q(a x+y)+1=0 \end{aligned}

Solve for q

q=2(ax+y)±4(ax+y)24(a2+1)2(a2+1)q=\frac{2(a x+y) \pm \sqrt{4(a x+y)^{2}-4\left(a^{2}+1\right)}}{2\left(a^{2}+1\right)}

Let's (ax+y)=u(a x+y)=u and (a2+1)=b=\left(a^{2}+1\right)=b= const

q=u±u2bbq=\frac{u \pm \sqrt{u^{2}-b}}{b}

p=aq=ab(u±u2b)p=a q=\frac{a}{b}\left(u \pm \sqrt{u^{2}-b}\right)

We know that

dz=pdx+qdy=aqdx+qdy=q(adx+dy)=qd(ax+y)=qdud z=p d x+q d y=a q d x+q d y=q(a d x+d y)=q d(a x+y)=q d u

Integrating


z=u±u2bbdu=u22b±1bu2bdu=z=\int \frac{u \pm \sqrt{u^{2}-b}}{b} d u=\frac{u^{2}}{2 b} \pm \frac{1}{b} \int \sqrt{u^{2}-b} d u=

=u22b±12uu2b12bln(u2b+u)+C==(ax+y)22b±12(ax+y)(ax+y)2b12bln((ax+y)2b+(ax+y))+C\begin{aligned} &\quad=\frac{u^{2}}{2 b} \pm \frac{1}{2} u \sqrt{u^{2}-b}-\frac{1}{2} b \ln \left(\sqrt{u^{2}-b}+u\right)+C= \\ &\quad=\frac{(a x+y)^{2}}{2 b} \pm \frac{1}{2}(a x+y) \sqrt{(a x+y)^{2}-b}-\frac{1}{2} b \ln \left(\sqrt{(a x+y)^{2}-b}+(a x+y)\right)+C \end{aligned}


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