Find the general solution of px + qy = 3z, p = (∂ z)/(∂ x)…

Find the general solution of \[ px + qy = 3z, \qquad p = \frac{\partial z}{\partial x},\quad q = \frac{\partial z}{\partial y}. \]

Step 1 · Characteristic equations

\[ \frac{dx}{x} = \frac{dy}{y} = \frac{dz}{3z}. \tag{1} \]

Step 2 · First integral

From $\dfrac{dx}{x} = \dfrac{dy}{y}$: $\ln x = \ln y + \text{const} \Rightarrow \dfrac{y}{x} = c_1$. (I)

Step 3 · Second integral

From $\dfrac{dx}{x} = \dfrac{dz}{3z}$: $3\ln x = \ln z + \text{const} \Rightarrow \dfrac{z}{x^3} = c_2$. (II)

Step 4 · General solution

The general solution is $\Phi(c_1,c_2)=0$, i.e. \[ z = x^3\,\Phi\!\left(\frac{y}{x}\right), \] where $\Phi$ is an arbitrary differentiable function.

Verification

Let $u=y/x$. Then $p = 3x^2\Phi(u) - xy\,\Phi'(u)$, $q = x^2\Phi'(u)$. Check $px+qy$: \begin{align*} px + qy &= x\!\left(3x^2\Phi - xy\,\Phi'\right) + y\cdot x^2\Phi' \\ &= 3x^3\Phi - x^2 y\,\Phi' + x^2 y\,\Phi' = 3x^3\Phi = 3z. \checkmark \end{align*}