Find the general solution of yp + xq = x - y, p = (∂ z)/(∂…

Find the general solution of \[ yp + xq = x - y, \qquad p = \frac{\partial z}{\partial x},\quad q = \frac{\partial z}{\partial y}. \]

Step 1 · Characteristic equations

\[ \frac{dx}{y} = \frac{dy}{x} = \frac{dz}{x-y}. \tag{1} \]

Step 2 · First integral

From $\dfrac{dx}{y}=\dfrac{dy}{x}$: $x\,dx=y\,dy \Rightarrow x^2-y^2=c_1$. (I)

Step 3 · Second integral

Take the combination with multipliers $(1,-1)$ on the first two ratios and set $v=x-y$, so $dv=dx-dy$ and $y-x=-v$: \[ \frac{dx-dy}{y-x} = \frac{dz}{x-y} \;\Longrightarrow\; \frac{dv}{-v} = \frac{dz}{v}. \] Cross-multiplying gives $v\,dv=-v\,dz$, which (dividing by $v\neq0$) simplifies to \[ dz = -dv = -(dx-dy) = dy-dx. \] Integrate: $z = y-x+c_2 \Rightarrow z-y+x=c_2$. (II)

Step 4 · General solution

The general solution is $c_2=\Phi(c_1)$: \[ z = y - x + \Phi\!\left(x^2 - y^2\right), \] where $\Phi$ is an arbitrary differentiable function.

Verification

Let $w=x^2-y^2$. Then $p=-1+2x\Phi'(w)$, $q=1-2y\Phi'(w)$. \begin{align*} yp+xq &= y(-1+2x\Phi') + x(1-2y\Phi') = -y+2xy\Phi'+x-2xy\Phi' = x-y. \checkmark \end{align*}