Problem 35
Medium First-Order PDEs

Find the general solution of cos y (∂ z)/(∂ x) + cos x (∂…

Find the general solution of \[ \cos y\,\frac{\partial z}{\partial x} + \cos x\,\frac{\partial z}{\partial y} = \cos x\cos y, \qquad p = \frac{\partial z}{\partial x},\quad q = \frac{\partial z}{\partial y}. \]

Explanation of Method

A Lagrange PDE with $P=\cos y$, $Q=\cos x$, $R=\cos x\cos y$. Both first integrals follow from direct, separable pairings of the characteristic ratios.

Step 1 · Characteristic equations

\[ \frac{dx}{\cos y} = \frac{dy}{\cos x} = \frac{dz}{\cos x\cos y}. \tag{1} \]

Step 2 · First integral

From $\dfrac{dx}{\cos y}=\dfrac{dy}{\cos x}$: $\cos x\,dx=\cos y\,dy$. Integrating: $\sin x - \sin y = c_1$. \tag{I}

Step 3 · Second integral

From the first and third ratios: $dz=\cos x\,dx$. Integrating: $z - \sin x = c_2$. \tag{II}

Step 4 · General solution

The general solution is $c_2=\Phi(c_1)$: \[ z = \sin x + \Phi(\sin x - \sin y), \] where $\Phi$ is an arbitrary differentiable function.


Verification

Let $u=\sin x-\sin y$. Then $p=\cos x(1+\Phi'(u))$, $q=-\cos y\,\Phi'(u)$. \begin{align*} \cos y\,p+\cos x\,q &= \cos x\cos y(1+\Phi') - \cos x\cos y\,\Phi' = \cos x\cos y. \checkmark \end{align*}

Result

The general solution is:

\[ \boxed{z = \sin x + \Phi(\sin x - \sin y)} \] where $\Phi$ is an arbitrary differentiable function.