Answer to Question #299376 in Differential Equations for Pankaj

Question

Solve (p + q)(px + qy) = 1, using Charpit’s method.

Answer 1

Solution:

The given equation is:

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

Thus the Charpit's auxiliary equations are:

${dp \over p(p+q)}={dq \over q(p+q)}= {dz \over -p(2px+qy+qx)-q(py+px+2qy)}={dx \over {-(2px+qy+qx)}}={dy \over {-(py+px+2qy)}}$

Taking the first two equations we get $p = aq$ for some constant $a$ . Putting it in original equation we get:

$q^2 = {1 \over (a+1)(ax+y)}$

and

$p^2 ={a^2 \over (a+1)(ax+y)}$ .

Thus $p= {a \over \sqrt{(a+1)(ax+y)}}$

$p= {a \over \sqrt{(a+1)(ax+y)}}$ and $q= {1 \over \sqrt{(a+1)(ax+y)}}$

Putting this values of $p, q$ in $dz = pdx + qdy$ we get:

$dz = {d(ax+y)\over \sqrt{(a+1)(ax+y)}}$

Thus integrating we get,

$z(a+1)^{\frac 12}=2(ax+y)^{\frac 12}+b$

where $a, b$ are constants.