Answer to Question #114852 in Differential Equations for Sheela John

Question

Solve the following
z(p^2+q^2)+px+qy=0

Answer 1

Given $(p^2+q^2)z+px+qy = 0$ _______________________(1)

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

Using Charpit's Method:

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

From last two factor, we get

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

So by integrating, we get $log(p) = log(q) - log(a) \implies q = ap$.

By putting value of $q$ in (1), we get

$p^2(1+a^2)z + px+apy = 0 \\ \implies p (1+a^2)z + (x+ay) = 0 \\ \implies p = -\frac{x+ay}{(1+a^2)z} \ and \ q = -\frac{a(x+ay)}{(1+a^2)z}$.

Finally, we have to solve: $dz = pdx+qdy$

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

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

So by integrating, we get

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

$\implies (1+a^2)z^2 = -x^2 - 2axy -a^2 y^2$ is the solution of given differential equation.