Question

P2-xp-q=0 solve by charpit method

Answer 1

Given differential equation is $p^2 - xp-q=0\Rightarrow p^2 = q + xp$ (1)

We can write it as $f(x,y,z,p,q) = q + xp - p^2 = 0$

$\frac{\partial f }{\partial x } = p$ , $\frac{\partial f }{\partial y } = 0$ , $\frac{\partial f }{\partial z } = 0$ , $\frac{\partial f }{\partial p} = x-2p$ , $\frac{\partial f }{\partial q } = 1$

Hence, equation will be

$\frac{dx}{-\frac{\partial f }{\partial p } } = \frac{dy}{-\frac{\partial f }{\partial q } } = \frac{dz}{-p\frac{\partial f }{\partial p }-q\frac{\partial f }{\partial q } } = \frac{dp}{\frac{\partial f }{\partial x }+p\frac{\partial f }{\partial z } } = \frac{dz}{\frac{\partial f }{\partial y }+q\frac{\partial f }{\partial z } }$

$\frac{dx}{2p-x } = \frac{dy}{-1 } = \frac{dz}{-p(x-2p)-q } = \frac{dp}{p} = \frac{dz}{0 }$

Taking $\frac{dy}{-1 } = \frac{dp}{p}$

Integrating it we get,

$p = ae^{-y}$

Then from given equation (1),

$q = -axe^{-y} + a^2e^{-2y}$

Now, $dz = pdx + qdy = ae^{-y}dx + (-axe^{-y}+a^2e^{-2y})dy$

Integrating it both sides, we get

$z = axe^{-y}-\frac{1}{2}a^2e^{-2y} + b$