Question

Find the complete integral using Charpit Method : 2x(q²z²+1) = pz

Answer 1

Given : $f(x,y,z,p,q)=2xq^2z^2+2x-pz$

Now we use Charpit's method, for which the following integrals are needed:

$\partial f/\partial x=2q^2z^2+2$

$\partial f/\partial y=0$

$\partial f /\partial /z=4xq^2z-p$

$\partial f/\partial p=-z$

$\partial f/\partial q=4xqz^2$

Clearly, we take the easiest two terms from Charpit's equation; we get;

$dx/z =dy/(-4xqz^2)$

Solving this we get;

$-\int 4xqzdx=\int dy$

$\implies -2qzx^2=y+a$

$\implies q=-(y+a)/(2x^2z)$ , where $a$ is a constant of integration

Putting this value of $q$ in $f(x,y,z,p,q)$ we get;

$pz=2xz^2[-(y+a)/2x^2z]^2+2x$

$\implies p=2x/z + [(y+a)^2/2x^3z]$

Now we put these obtained values in : $pdx+qdy=dz$

$dz=(2x/z+[(y+a)^2/2x^3z])dx - ((y+a)/2x^2z$$)dy$

$zdz=2xdx+1/4d((y+a)^2/x^2)$

Integrating we get:

$z^2/2=x^2+[(y+a)^2/4x^2]+b$ (Answer)

where $a$ and $b$ are constants of integration.