Answer to Question #183514 in Differential Equations for Priya

Question

p= (z+qy)² how to solve it by Jacobi 's method

Answer 1

:Here the given partial differential equation is

$f(x, y, z, p, q) = p − (qy + z)^2 = 0, ~~~~~~~-(1)$

Now, the auxiliary equations are

$\dfrac{dp}{\frac{∂f}{∂x} +p \frac{∂f}{∂z}}=\dfrac{dp}{\frac{∂f}{∂y} + q\frac{∂f}{∂z}}=\dfrac{dz}{−p\frac{∂f}{∂p} − q\frac{∂f}{∂q}}=\dfrac{dx}{−\frac{∂f}{∂p}}=\dfrac{dy}{−\frac{∂f}{∂q}}$

or,

$\dfrac{dp}{2p(qy + z)}=\dfrac{dq}{2q(qy + z)}=\dfrac{dz}{(−p)(−1) − q.2(qy + z)y}=\dfrac{dx}{−(−1)} =\dfrac{dy}{−2y(qy + z)}$

Taking the 1st and 5th ratios, we get

$\dfrac{dp}{p}+\dfrac{dy}{y}= 0,$

Integrating, $log p + log y = log a$ , which gives, $p =\dfrac{a}{y}.$

Substitute in (1), we have,$\dfrac{a}{y} = (qy + z)^2$ , which gives, $q =\sqrt{\dfrac{a}{y}}-\dfrac{z}{y}$

Substituting these value of p and q in $dz = p dx + q dy$ , we get,

$dz =\dfrac{a}{y}dx + (\sqrt{\dfrac{a}{y}}−\dfrac{z}{y}) dy$

or

$y dz + z dy = a dx +\sqrt{\dfrac{a}{y}}dy$

After integration, we get, $yz = ax +2\sqrt{(ay)}+ b$ , which is a complete integral

with a and b are arbitrary constants.