Solve the system of partial differential equations (∂ z)/(∂…
Solve the system of partial differential equations \[ \frac{\partial z}{\partial x} = 2yz - z^2, \qquad \frac{\partial z}{\partial y} = xz. \]
Explanation of Method
Both equations are Bernoulli-type in $z$; the substitution $w=1/z$ linearises them. Checking compatibility of the resulting linear system reveals a contradiction, so the original system has no solution.
Step 1 · Linearise via $w=1/z$
With $w=1/z$ (so $z_x=-w_x/w^2$, $z_y=-w_y/w^2$): \[ w_x + 2yw = 1, \tag{1} \qquad w_y = -xw. \tag{2} \]
Step 2 · Compatibility of the linear system
From (1): $w_x=1-2yw$. Requiring $\partial_y(w_x)=\partial_x(w_y)$: \[ \frac{\partial}{\partial y}(1-2yw) = -2w-2yw_y = -2w+2xyw, \] \[ \frac{\partial}{\partial x}(-xw) = -w-xw_x = -w-x(1-2yw) = -w-x+2xyw. \] Equating: $-2w+2xyw=-w-x+2xyw \Rightarrow -w=-x \Rightarrow w=x$, i.e. $z=1/x$.
Step 3 · Test $z=1/x$ against the original equations
If $z=1/x$: $z_x=-1/x^2$, while $2yz-z^2=2y/x-1/x^2$. Equality would require $2y/x=0$ for all $y$, which is false. So the compatibility condition again forces a function that fails to satisfy the equations.
Step 4 · Confirm via direct integration
Solving (1) directly as a linear ODE in $x$ gives $w=\tfrac1{2y}+\varphi(y)e^{-2yx}$; substituting into (2) forces both $\varphi\equiv0$ (from the coefficient of $e^{-2yx}$) and $\tfrac1{2y^2}=\tfrac{x}{2y}$ for all $x$ (from the remaining term) — two conditions that cannot hold simultaneously. This confirms the system is overdetermined.
Result
Conclusion: