Find the general solution of x (∂ z)/(∂ x) + y (∂ z)/(∂ y) +…

Find the general solution of \[ x\,\frac{\partial z}{\partial x} + y\,\frac{\partial z}{\partial y} + t\,\frac{\partial z}{\partial t} = xyt, \] where $z = z(x, y, t)$.

Step 1 · Characteristic equations

\[ \frac{dx}{x} = \frac{dy}{y} = \frac{dt}{t} = \frac{dz}{xyt}. \tag{1} \]

Step 2 · First and second integrals

From $\dfrac{dx}{x}=\dfrac{dy}{y}$: $\dfrac{y}{x}=c_1$. (I) From $\dfrac{dx}{x}=\dfrac{dt}{t}$: $\dfrac{t}{x}=c_2$. (II)

Step 3 · Third integral

From $\dfrac{dx}{x}=\dfrac{dz}{xyt}$: along characteristics $y=c_1x$, $t=c_2x$, so \[ \frac{dx}{x} = \frac{dz}{c_1 c_2 x^3} \;\Longrightarrow\; dz = c_1c_2x^2\,dx \;\Longrightarrow\; z=\frac{c_1c_2x^3}{3}+c_3. \] Substituting back $c_1=y/x$, $c_2=t/x$: $z=\dfrac{xyt}{3}+c_3$, so \[ c_3 = z - \frac{xyt}{3}. \tag{III} \]

Step 4 · General solution

The general solution is an arbitrary functional relation among $c_1,c_2,c_3$: \[ z - \frac{xyt}{3} = \Phi\!\left(\frac{y}{x},\,\frac{t}{x}\right). \]

Verification

Let $u=y/x$, $v=t/x$. Computing $z_x,z_y,z_t$ and substituting: \begin{align*} x z_x + y z_y + t z_t &= xyt + \Phi_u\!\left(-\frac{y}{x} + \frac{y}{x}\right) + \Phi_v\!\left(-\frac{t}{x} + \frac{t}{x}\right) = xyt. \checkmark \end{align*}