$\cos(x+y)\frac{\partial z}{\partial x}+\sin(x+y)\frac{\partial z}{\partial y}=z+\frac{1}{z}$

Question

Having difficulty in solving the following partial differential equation: $$\cos(x+y)\frac{\partial z}{\partial x}+\sin(x+y)\frac{\partial z}{\partial y}=z+\frac{1}{z}.$$

Will it be easier if we substitute $x+y$ as $s$ and proceed?

Answer 1

Let $\begin{cases}p=x+y\\q=x-y\end{cases}$ ,

Then $\dfrac{\partial z}{\partial x}=\dfrac{\partial z}{\partial p}\dfrac{\partial p}{\partial x}+\dfrac{\partial z}{\partial q}\dfrac{\partial q}{\partial x}=\dfrac{\partial z}{\partial p}+\dfrac{\partial z}{\partial q}$

$\dfrac{\partial z}{\partial y}=\dfrac{\partial z}{\partial p}\dfrac{\partial p}{\partial y}+\dfrac{\partial z}{\partial q}\dfrac{\partial q}{\partial y}=\dfrac{\partial z}{\partial p}-\dfrac{\partial z}{\partial q}$

$\therefore\cos p\left(\dfrac{\partial z}{\partial p}+\dfrac{\partial z}{\partial q}\right)+\sin p\left(\dfrac{\partial z}{\partial p}-\dfrac{\partial z}{\partial q}\right)=z+\dfrac{1}{z}$

$(\sin p+\cos p)\dfrac{\partial z}{\partial p}+(\cos p-\sin p)\dfrac{\partial z}{\partial q}=\dfrac{z^2+1}{z}$

$\dfrac{\partial z}{\partial p}+\dfrac{\cos p-\sin p}{\sin p+\cos p}\dfrac{\partial z}{\partial q}=\dfrac{z^2+1}{z(\sin p+\cos p)}$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dp}{dt}=1$ , letting $p(0)=0$ , we have $p=t$

$\dfrac{dq}{dt}=\dfrac{\cos p-\sin p}{\sin p+\cos p}=\dfrac{\cos t-\sin t}{\sin t+\cos t}$ , letting $q(0)=q_0$ , we have $q=q_0+\ln(\sin t+\cos t)=q_0+\ln(\sin p+\cos p)$

$\dfrac{dz}{dt}=\dfrac{z^2+1}{z(\sin p+\cos p)}=\dfrac{z^2+1}{z(\sin t+\cos t)}$ , we have $z=\pm\sqrt{f(q_0)\tan^\sqrt2\left(\dfrac{t}{2}+\dfrac{\pi}{8}\right)-1}=\pm\sqrt{f(q-\ln(\sin p+\cos p))\tan^\sqrt2\left(\dfrac{p}{2}+\dfrac{\pi}{8}\right)-1}=\pm\sqrt{f(x-y-\ln(\sin(x+y)+\cos(x+y)))\tan^\sqrt2\left(\dfrac{x+y}{2}+\dfrac{\pi}{8}\right)-1}$