Find the general solution of sin^2 x (∂ z)/(∂ x) + tan z (∂…
Find the general solution of \[ \sin^2 x\,\frac{\partial z}{\partial x} + \tan z\,\frac{\partial z}{\partial y} = \cos^2 z, \qquad p = \frac{\partial z}{\partial x},\quad q = \frac{\partial z}{\partial y}. \]
Explanation of Method
A Lagrange PDE with $P=\sin^2x$, $Q=\tan z$, $R=\cos^2z$. The first integral is separable using the first and third ratios; the second uses the second and third ratios, integrated with the substitution $u=\cos z$.
Step 1 · Characteristic equations
\[ \frac{dx}{\sin^2 x} = \frac{dy}{\tan z} = \frac{dz}{\cos^2 z}. \tag{1} \]
Step 2 · First integral
From $\dfrac{dx}{\sin^2x}=\dfrac{dz}{\cos^2z}$: $\csc^2x\,dx=\sec^2z\,dz$. Integrating: $-\cot x=\tan z+c_1$, so \[ \tan z + \cot x = c_1. \tag{I} \]
Step 3 · Second integral
From $\dfrac{dy}{\tan z}=\dfrac{dz}{\cos^2z}$: $dy=\dfrac{\sin z}{\cos^3z}\,dz$. With $u=\cos z$, $\displaystyle\int\frac{\sin z}{\cos^3z}\,dz=\frac{1}{2\cos^2z}+\text{const}$, so \[ y - \frac{\sec^2 z}{2} = c_2. \tag{II} \]
Step 4 · General solution
The general solution is $c_2=\Phi(c_1)$: \[ y - \frac{\sec^2 z}{2} = \Phi\!\left(\tan z + \cot x\right), \] where $\Phi$ is an arbitrary differentiable function.
Verification
First integral: $c_{1x}=-\csc^2x$, $c_{1z}=\sec^2z$: \[ Pc_{1x}+Rc_{1z} = \sin^2x(-\csc^2x)+\cos^2z\sec^2z = -1+1=0. \checkmark \] Second integral: $c_{2y}=1$, $c_{2z}=-\sec^2z\tan z$: \[ Qc_{2y}+Rc_{2z} = \tan z - \cos^2z\sec^2z\tan z = \tan z-\tan z = 0. \checkmark \] Since both first integrals satisfy the characteristic condition, every surface $y-\sec^2z/2=\Phi(\tan z+\cot x)$ solves the PDE. $\checkmark$
Result
The general solution is: