To Solve a linear PDE of first order

Question

To Solve: $\displaystyle \cos(x+y)\frac{\partial z}{\partial x}+\sin(x+y)\frac{\partial z}{\partial y}=z$

My attempt:

Forming the subsidiary equations: $\displaystyle \frac{dx}{\cos(x+y)}=\frac{dy}{\sin(x+y)}=\frac{dz}{z}$

I was hoping to use the method of multipliers or method of grouping, but can't think of anything here..

The given answer is: $\displaystyle[\cos(x+y)+\sin(x+y)]c^{y-x}=\phi\left[z^{\sqrt 2}\tan\left(\frac{x+y}{2}+\frac{\pi}{8}\right )\right]$

How did we get here ?

Answer 1

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

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

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

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

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

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

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 $q=q_0+\ln(\sin t+\cos t)=q_0+\ln(\sin p+\cos p)$

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

Answer 2

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

Setting $z(x,y)=w(u)$, $u=x+y$, we have $$\frac{\partial z(x,y)}{\partial x}=\frac{dw(u)}{du}\frac{\partial u}{\partial x}=\frac{dw(u)}{du}$$

$$\frac{\partial z(x,y)}{\partial y}=\frac{dw(u)}{du}\frac{\partial u}{\partial y}=\frac{dw(u)}{du}$$

So the original equation becomes

$$ (\cos u+\sin u)\frac{d w(u)}{d u}=w(u)$$

Setting $$\cos u=\frac{1-t^2}{1+t^2}, \sin u=\frac{2t}{1+t^2}, t=\tan(u/2), w(u)=f(t)$$, We obtain: $$ du=\frac{2dt}{1+t^2}, dw(u)=df(t)$$

Thus the equation above becomes: $$ (1+2t-t^2)\frac{df(t)}{dt}=2f(t)$$

The solution is given by:

$$\ln f(t)=\ln c_1+2^{-3/2}\ln \frac{\sqrt{2}-1+t}{\sqrt{2}+1-t}$$

or

$$f(t)=c_1\left(\frac{\sqrt{2}-1+t}{\sqrt{2}+1-t}\right)^{2^{-3/2}}$$

I have no idea how to go from here to the final expression you have.

-mike