Find the general solution of (∂ z)/(∂ x) - a (∂ z)/(∂ y) =…

Find the general solution of \[ \frac{\partial z}{\partial x} - a\,\frac{\partial z}{\partial y} = e^{mx}\cos py, \] where $a$, $m$, $p$ are real constants.

Step 1 · Characteristic equations

\[ \frac{dx}{1} = \frac{dy}{-a} = \frac{dz}{e^{mx}\cos py}. \tag{1} \]

Step 2 · First integral

From $dx=-dy/a$: $y+ax=c_1$. \tag{I}

Step 3 · Second integral

From $dz=e^{mx}\cos py\,dx$ with $y=c_1-ax$ along the characteristic: \[ dz = e^{mx}\cos\bigl(p(c_1-ax)\bigr)\,dx. \] Expanding $\cos(pc_1-pax)$ and using $\int e^{\alpha x}\cos(\beta x)\,dx = \dfrac{e^{\alpha x}(\alpha\cos\beta x+\beta\sin\beta x)}{\alpha^2+\beta^2}$ with $\alpha=m$, $\beta=pa$, the angle-addition identities collapse the result to \[ z = \frac{e^{mx}(m\cos p(c_1-ax) - pa\sin p(c_1-ax))}{m^2+p^2a^2} + c_2. \] Since $c_1-ax=y$: \[ c_2 = z - \frac{e^{mx}(m\cos py - pa\sin py)}{m^2+p^2a^2}. \tag{II} \]

Step 4 · General solution

The general solution $c_2=\Phi(c_1)$ is: \[ z = \frac{e^{mx}(m\cos py - pa\sin py)}{m^2+p^2a^2} + \Phi(y+ax), \] where $\Phi$ is an arbitrary differentiable function.

Verification

Let $f=\dfrac{e^{mx}(m\cos py-pa\sin py)}{m^2+p^2a^2}$. Then $f_x = \dfrac{me^{mx}(m\cos py-pa\sin py)}{m^2+p^2a^2}$, $f_y = \dfrac{-pe^{mx}(m\sin py+pa\cos py)}{m^2+p^2a^2}$. Checking $f_x-af_y$, the $\sin py$ and $\cos py$ cross terms cancel, leaving \[ f_x-af_y = \frac{e^{mx}(m^2+p^2a^2)\cos py}{m^2+p^2a^2} = e^{mx}\cos py. \checkmark \] For the homogeneous part, with $\xi=y+ax$: $\Phi_x-a\Phi_y=a\Phi'(\xi)-a\Phi'(\xi)=0$. $\checkmark$ Hence $z_x-az_y=e^{mx}\cos py$. $\checkmark$