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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: $displaystyle[cos(x+y)+sin(x+y)]c^{y-x}=phileft[z^{sqrt 2}tanleft(frac{x+y}{2}+frac{pi}{8}right )right]$

How did we get here ?

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

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

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

$thereforecos pleft(dfrac{partial z}{partial p}+dfrac{partial z}{partial q}right)+sin pleft(dfrac{partial z}{partial p}-dfrac{partial z}{partial q}right)=z$

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

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

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

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

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

$$f(t)=c_1left(frac{sqrt{2}-1+t}{sqrt{2}+1-t}right)^{2^{-3/2}}$$

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

-mike