Question

find the compute integral of p²x+q²y=z;using charpits method

Answer 1

Let's solve the problem

$f = p^2x+q^2y-z = 0$

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

$\large\frac{2pxdp+p^2xdx}{2px(-p+p^2)+p^2(-2px)}= \large\frac{2qydq+q^2dy}{2qy(-q+q^2)+q^2(-2qy)}$

$\large\frac{d(p^2x)}{-2p^2x} = \large\frac{2(q^2y)}{-2qy}$

$\large\frac{}{}$ $\large\frac{d(p^2x)}{p^2x} = \large\frac{d(q^2y)}{q^2y}$

$log(p^2x)=log(q^2)+loga$

$p^2x = q^2ya$

$aq^2y+q^2y = z \space\space \space\space \space\space \space\space q^2y(1+a)=z$

$q = \large[\large\frac{z}{(1+a)y}]^{\frac{1}{2}}$

$p^2 = \large\frac{q^2ya}{x}$

$p = q\large(\frac{ya}{x})^\frac{1}{2} = [\frac{2a}{(1+a)x}]^{\frac{1}{2}}$

$dz = pdx+qdy$

$dz = \large\{\frac{za}{(1+a)x}\}$ $dx + \large\{\frac{za}{(1+a)y}\}$ $dy$

Answer: $(1+a)^{1/2}\sqrt{z} = \sqrt{a}\sqrt{x}+\sqrt{y} + b$