Answer to Question #149882 in Differential Equations for Nikhil Singh

Question

Apply Charpit's method to solve the equation z-px-qy=p^2+q^2

Answer 1

Given differential equation is $z-px-qy-p^2-q^2=0$

Then, $f(x,y,z,p,q) = z-px-qy-p^2-q^2=0$

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

Then,

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

Putting values, we get,

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

$\frac{dx}{x+2p} = \frac{dy}{y+2q}=\frac{dz}{px+qy+2p^2+2q^2}=\frac{dp}{0}=\frac{dq}{0}$

Taking second last term, $dp = 0 \implies p = a$ (1)

taking last term, $dq=0 \implies q=b$ (2)

Then, $dz = pdx+qdy = adx+bdy$

integrating both sides, $z = ax+by+c$ (3)

Now, equation will be,

$ax+by+c = ax+by+a^2+b^2$

$\implies c = a^2+b^2$

Hence, $z = ax+by+a^2+b^2$