Question

find a complete integral of z=px+qy+p^2+q^2​

Answer 1

Answer:

The given equation isF=z−px−qy−p2−q2=0.

Charpit's auxiliary equations aredpFx+pFz=dqFy+qFz=dz−pFp−qFq=dx−Fp=dy−Fq⟹dp−p+p(1)=dq−q+q(1)=dz−p(−x−2p)−q(−y−2q)=dx−x−2p=dy−y−2q⟹dp0=dq0=dzpx+qy+2p2+2q2=dx−x−2p=dy−y−2q

First fraction implies, dp=0⟹p=a.

Similarly, let q=b.

Now, dz=pdx+qdy⟹dz=adx+bdy⟹z=ax+by+c.

Putting the value of z in the given equation,

ax+by+c=ax+by+a2+b2⟹c=a2+b2

Therefore, z=ax+by+a2+b2.

This is the complete solution.

Answer 2

Let $F=px+qy+p^2+q^2=0$. Then by Charpit's auxiliary equations

$\dfrac{dp}{F_x+pF_z}=\dfrac{dq}{F_y+qF_z}=\dfrac{dz}{-pF_p-qF_q}=\dfrac{dx}{-F_p}=\dfrac{dy}{-F_q}$

we have

$\dfrac{dp}{0}=\dfrac{dq}{0}=\dfrac{dz}{-p(x+2p)-q(y+2q)}=\dfrac{dx}{-(x+2p)}=\dfrac{dy}{-(y+2q)}$ .

This implies $p=a$=constant. Putting this in given equation, $q=\dfrac{-y\pm \sqrt{y^2-4(a^2+ax-z)}}{2}$.