Solve z=pq+1 subject to z=2x+1 on the line y=2

Solve $z=pq+1$ subject to $z=2x+1$ on the line $y=2$.

Step 1 · Set up Charpit's equations

With $F_x=F_y=0$, $F_z=1$, $F_p=-q$, $F_q=-p$: \[ \frac{dx}{dt}=-q,\quad\frac{dy}{dt}=-p,\quad\frac{dz}{dt}=-2pq=-2(z-1), \quad\frac{dp}{dt}=-p,\quad\frac{dq}{dt}=-q. \]

Step 2 · Determine the initial strip

On $x=x_0$, $y=2$, $z=2x_0+1$, the strip condition ($dy=0$ along the curve) gives $p_0=dz/dx_0=2$; the PDE itself then gives $2x_0+1=2q_0+1\Rightarrow q_0=x_0$.

Step 3 · Integrate the characteristics

\[ p=2e^{-t},\qquad q=x_0e^{-t}, \] \[ x=x_0e^{-t}\ (\text{from } \dot x=-q,\ x(0)=x_0),\qquad y=2e^{-t}\ (\text{from }\dot y=-p,\ y(0)=2), \] \[ z-1=(z_0-1)e^{-2t}=2x_0e^{-2t}\ (\text{from }\dot z=-2(z-1)). \]

Step 4 · Eliminate $t,x_0$

From $x,y$: $e^{-t}=y/2$, $x_0=x/e^{-t}=2x/y$. Substitute into $z=1+2x_0e^{-2t}$: \[ z = 1+2\cdot\frac{2x}{y}\cdot\frac{y^2}{4} = 1+xy. \]

Verification

$z=xy+1$: $p=y$, $q=x$, so $pq+1=xy+1=z$. \checkmark At $y=2$: $z=2x+1$. \checkmark