Solve z=px+qy+2pq subject to z=4y+12 on x=-4
Solve $z=px+qy+2pq$ subject to $z=4y+12$ on $x=-4$.
Explanation of Method
For a Clairaut equation $p,q$ are constant along every characteristic (Charpit gives $\dot p=\dot q=0$), so the solution is a plane; determine its slopes from the initial strip and integrate.
Step 1 · Charpit's equations confirm $p,q$ constant
With $F=px+qy+2pq-z$: $F_x=p$, $F_z=-1$, so $\dot p=-(F_x+pF_z)=-(p-p)=0$; similarly $\dot q=0$.
Step 2 · Determine the initial strip
On $x=-4$, $y=y_0$, $z=4y_0+12$: $dx=0$ gives $q_0=dz/dy_0=4$. Substituting $x=-4$, $q_0=4$, $z=4y_0+12$ into the PDE: \[ 4y_0+12 = -4p_0+4y_0+8p_0 = 4p_0+4y_0 \implies p_0=3. \] Since $p,q$ are constant everywhere: $p=3$, $q=4$.
Step 3 · Integrate
\[ z = 3x+4y+C. \] At $x=-4$: $4y+12=-12+4y+C \implies C=24$.
Verification
$z=3x+4y+24$: $p=3$, $q=4$, $px+qy+2pq=3x+4y+24=z$. \checkmark At $x=-4$: $z=4y+12$. \checkmark
Result
The solution of the Cauchy problem is: