Solve 2z=pq-3xy subject to z=15y on x=5
Solve $2z=pq-3xy$ subject to $z=15y$ on $x=5$.
Explanation of Method
Form Charpit's characteristic system for $F=pq-3xy-2z=0$; determine the initial strip from the given curve, reduce the resulting linear ODE system for $x(t),y(t)$ to a single second-order equation, and eliminate the parameters.
Step 1 · Set up Charpit's equations
With $F_x=-3y$, $F_y=-3x$, $F_z=-2$, $F_p=q$, $F_q=p$: \[ \dot x=q,\ \dot y=p,\ \dot z=2pq,\ \dot p=3y+2p,\ \dot q=3x+2q. \]
Step 2 · Determine the initial strip
On $x=5$, $y=y_0$, $z=15y_0$: the strip condition ($dx=0$) gives $q_0=dz/dy_0=15$; the PDE then gives $30y_0=15p_0-15y_0 \Rightarrow p_0=3y_0$.
Step 3 · Solve the characteristic ODEs
Differentiating $\dot x=q$ and using $\dot q=3x+2q$ gives $\ddot x-2\dot x-3x=0$, with roots $3,-1$. Using $x(0)=5$, $q(0)=15$ fixes the solution: \[ x=5e^{3t},\qquad q=15e^{3t}. \] The same equation governs $y$ (via $\dot y=p$, $\dot p=3y+2p$); using $y(0)=y_0$, $p(0)=3y_0$: \[ y=y_0e^{3t},\qquad p=3y_0e^{3t}. \]
Step 4 · Integrate for $z$ and eliminate $t,y_0$
\[ \dot z=2pq=90y_0e^{6t} \implies z=15y_0e^{6t}\ (\text{using }z(0)=15y_0). \] From $x=5e^{3t}$: $e^{6t}=x^2/25$. From $y=y_0e^{3t}$: $y_0=5y/x$. Substituting: \[ z = 15\cdot\frac{5y}{x}\cdot\frac{x^2}{25} = 3xy. \]
Verification
$z=3xy$: $p=3y$, $q=3x$, so $pq-3xy=9xy-3xy=6xy=2z$. \checkmark At $x=5$: $z=15y$. \checkmark
Result
The solution of the Cauchy problem is: