Question

yp+xq+pq=0

Answer 1

Given the partial differential equation:

$$yp+xq+pq=0$$

The given equation can be written like:

$$yp + xq = – pq \implies \dfrac{y}{q}+\dfrac{x}{p} = -1$$

OR

$$\Big(\dfrac{y}{q} \Big) = \Big(-1-\dfrac{x}{p} \Big) = a\quad \text{(say)}$$

which belong to the category $f_1(x,p) = f_2(y,q)$ where:

$$\dfrac{y}{q} =a \implies q = \dfrac{y}{a}; \qquad \text{and}\\ -1 - \dfrac{x}{p} = a \implies p = \Big(\dfrac{x}{-1-a} \Big)$$

for $z(x,y):$

$$dz = \dfrac{\partial z}{\partial x} dx + \dfrac{\partial z}{\partial y} dy\\$$

$$dz =\Big(\dfrac{x}{-1-a} \Big)dx+\dfrac{y}{a}dy$$

Integrating through, we have:

$$z = \dfrac{x^2}{2(-1-a)}+\dfrac{y^2}{2a}+c\\ \implies 2z = - \dfrac{x^2}{(-1-a)}+\dfrac{y^2}{a}+b\\$$

where $b=2c$