Question

Solve Z(p^2+q^2)+px+qy=0

Answer 1

Let f=z($p^2+q^2)+px+qy$

Using charpit's method,

$\frac{dz}{-p\frac{df}{dp}-q\frac{df}{dq}}$ =$\frac{dp}{\frac{df}{dx}+p\frac{df}{dz}}$ =$\frac{dq}{\frac{df}{dy}+q\frac{df}{dz}}=\frac{dx}{\frac{-df}{dp}}=\frac{dy}{\frac{-df}{dq}}$

From second and third factor we have

$\frac{dp}{\frac{df}{dx}+p\frac{df}{dz}}=\frac{dq}{\frac{df}{dy}+q\frac{df}{dz}}$

$\frac{dp}{p+p(p^2+q^2)}=\frac{dq}{q+q(p^2+q^2)}$

$\frac{dp}{p}=\frac{dq}{q}$

So by integrating ,we get

$log(p)=log(q)-log(a)$

q=pa

Putting the value of p in given equation,

$p^2(1+a^2)z+px+apy=0$

$p(1+a^2)z+(x+ay)=0$

so p=$-\frac{x+ay}{z(1+a^2)}$ so q=$-\frac{a(x+ay)}{(1+a^2)z}$

now we have to solve

dz=pdx+qdy

$dz=-\frac{x+ay}{(1+a^2)z}dx-\frac{a(x+ay)}{1+a^2)z}dy$

$(1+a^2)zdz=-xdx-a(ydx+xdy)-a^2ydy$

On integrating the above equation, we get

$(1+a^2)\frac{z^2}{2}=-\frac{x^2}{2}-axy-a^2\frac{y^2}{2}$

$(1+a^2)z=-x^2-2axy-a^2y^2$

This is the solution.