Find the complete integral of z^3 = pq^2
Find the complete integral of \[ z^3 = pq^2. \]
Explanation of Method
Since $x,y$ are absent except through $p,q$, look for a solution depending only on the combination $\xi=ax+y$: $z=f(\xi)$, so $p=af'$, $q=f'$. This reduces the PDE to a separable first-order ODE for $f(\xi)$.
Step 1 · Substitute the travelling-wave ansatz
With $z=f(ax+y)$, $p=af'$, $q=f'$: \[ f^3 = af'\cdot(f')^2 = a(f')^3. \]
Step 2 · Solve the resulting ODE
Taking the cube root: $f=a^{1/3}f'$, so \[ \frac{df}{f} = \frac{d\xi}{a^{1/3}} \implies f = e^{\xi/a^{1/3}} \] (absorbing the multiplicative integration constant into a later additive shift of $\xi$).
Step 3 · Write the complete integral
With $\xi=ax+y+b_0$ and relabelling $a\to a^3$ so the exponent is rational, and absorbing constants into $b$: \[ z = e^{\,a^2x+y/a+b}. \]
Verification
$p=a^2z$, $q=z/a$, so \[ pq^2 = a^2z\cdot\frac{z^2}{a^2} = z^3. \checkmark \]
Result
The complete integral is: