Find the general solution of yz p - xz q = e^z, p = (∂ z)/(∂…
Find the general solution of \[ yz\,p - xz\,q = e^z, \qquad p = \frac{\partial z}{\partial x},\quad q = \frac{\partial z}{\partial y}. \]
Explanation of Method
A Lagrange PDE with $P=yz$, $Q=-xz$, $R=e^z$. The first integral follows once $z$ cancels between the first two ratios; the second uses multipliers $(y,-x,0)$, which produces the polar angle $\arctan(y/x)$, reducing the equation to a separable ODE in that angle and $z$.
Step 1 · Characteristic equations
\[ \frac{dx}{yz} = \frac{dy}{-xz} = \frac{dz}{e^z}. \tag{1} \]
Step 2 · First integral
From $\dfrac{dx}{yz}=\dfrac{dy}{-xz}$, the factor $z$ cancels: $-x\,dx=y\,dy$. Integrating: $x^2+y^2=c_1$. \tag{I}
Step 3 · Second integral
Apply multipliers $(y,-x,0)$: \[ \frac{y\,dx-x\,dy}{y^2z+x^2z} = \frac{y\,dx-x\,dy}{z(x^2+y^2)} = \frac{dz}{e^z}. \] Since $d(\arctan(y/x)) = (x\,dy-y\,dx)/(x^2+y^2)$, the numerator is $-(x^2+y^2)\,d(\arctan(y/x))$, so \[ \frac{-d(\arctan(y/x))}{z} = \frac{dz}{e^z} \;\Longrightarrow\; -e^z\,d\!\left(\arctan\frac{y}{x}\right) = z\,dz. \] Separating and integrating (by parts, $\int ze^{-z}dz=-(z+1)e^{-z}$): \[ d\!\left(\arctan\frac{y}{x}\right) = -ze^{-z}\,dz \;\Longrightarrow\; \arctan\frac{y}{x} = (z+1)e^{-z} + c_2. \tag{II} \]
Step 4 · General solution
The general solution is $c_2=\Phi(c_1)$: \[ \arctan\frac{y}{x} - (z+1)e^{-z} = \Phi\!\left(x^2+y^2\right), \] where $\Phi$ is an arbitrary differentiable function.
Verification
First integral: $yz\cdot2x+(-xz)\cdot2y=0. \checkmark$
Second integral: with $F=\arctan(y/x)-(z+1)e^{-z}$, $F_x=-y/(x^2+y^2)$,
$F_y=x/(x^2+y^2)$, $F_z=ze^{-z}$:
\[ PF_x+QF_y+RF_z = \frac{-y^2z-x^2z}{x^2+y^2}+e^z\cdot ze^{-z} = -z+z=0. \checkmark \]
PDE check: implicit differentiation of $\arctan(y/x)-(z+1)e^{-z}=\Phi(x^2+y^2)$
with respect to $x$ and $y$, eliminating $\Phi'$ between the two relations, gives
\[ ze^{-z}(yp-xq)=1 \;\Longrightarrow\; yzp-xzq=e^z. \checkmark \]
Result
The general solution is: