General solutions for a PDE

Question

Given PDE : $$\cos(x+y) p + \sin(x+y) q = z ; $$ Find its General solution.
Solution:
Lagrange's Auxillary equations: $$\frac{dx}{\cos(x+y)} = \frac{dy}{\sin(x+y)} = \frac{dz}{z}\qquad .....{\rm eqn}.1 $$ Using the first two pairs derived : $$\frac{dy}{dx}= \tan(x+y) \qquad .....{\rm eqn}.2 $$
Using multipliers $(1,1,0)$ and $(1,-1,0)$ : $$\frac{dx+dy}{\cos(x+y) +\sin(x+y)}= \frac{dx-dy}{\cos(x+y) -\sin(x+y)} \qquad .....{\rm eqn}.3$$
Solving eqn.3 by integrating both sides, gives me a solution $$c_1=e^{y-x}(\cos(x+y)+\sin(x+y)) \qquad ....{\rm solution}.1$$
For second solution, Using eqn.1 and eqn.3 : $$\frac{dz}{z} =\frac{dx+dy}{\cos(x+y) +\sin(x+y)} $$
$$\implies\frac{dz}{z} = \frac{dx(1+\frac{dy}{dx})}{\cos(x+y)(1+\tan(x+y))}$$ From eqn.(2), $$\implies\frac{dz}{z} = \frac{dx(1+{\tan(x+y))}}{\cos(x+y)(1+\tan(x+y))}$$ $$\implies\frac{dz}{z} = \frac{dx}{\cos(x+y)}$$ Integrating both sides, gives me $$\implies\ln{z} = \ln|\sec(x+y)+\tan(x+y)|+ \ln{c_2}$$ $$\implies c_2= \frac{\sec(x+y)+\tan(x+y)}{z}\qquad .....{\rm solution}.2 $$
does $$\phi( e^{y-z}(\cos(x+y)+\sin(x+y)),\frac{\sec(x+y)+\tan(x+y)}{z})=0$$ stand as a valid solution for given PDE $\cos(x+y) p + \sin(x+y) q = z $?

Answer 1

Thanks to the inputs from @JJacquelin && @Gonçalo, posting the updated solution for above PDE.

For second solution, Using eqn.1 and eqn.3 : $$\implies\frac{dz}{z} =\frac{dx+dy}{\cos(x+y) +\sin(x+y)} $$
$$\implies\frac{dz}{z} =\frac{d(x+y)}{\cos(x+y) +\sin(x+y)} $$
let,$\quad x+y = t$
$$\implies\frac{dz}{z} =\frac{d(t)}{\cos(t) +\sin(t)} $$
$$\implies\sqrt{2}.\frac{dz}{z} = \frac{dt}{\sin(\frac{\pi}{4}).\cos(t)+(\cos(\frac{\pi}{4}).\sin(t))}$$ $$\implies\sqrt{2}.\frac{dz}{z} = \frac{dt}{sin(\frac{\pi}{4}+t))}$$ Integrating both sides, $$\implies\sqrt{2}.\frac{dz}{z} = \frac{dt}{sin(\frac{\pi}{4}+t))}$$ $$\implies\sqrt{2}\ln|z| = \ln|\tan(\frac{1}{2}.(\frac{\pi}{4}+(x+y))|+ \ln{c_2}$$ $$\implies\sqrt{2}\ln|z| = \ln|{c_2}.\tan(\frac{1}{2}.(\frac{\pi}{4}+(x+y))|$$ Taking anti-log both sides, $$\implies {c_2} = z^\sqrt{2}.\cot(\frac{1}{2}.(x+y+\frac{\pi}{4})) \qquad ....solution .02$$ $$\phi({c_1}, {c_2})= 0 $$ is a general solution of given PDE.

Answer 2

With the change of variables $(\xi,\eta)=(x+y,y)$, we may rewrite the PDE as $$ (\sin\xi + \cos\xi)z_{\xi}+(\sin\xi) z_{\eta}=z, \tag{1} $$ and the characteristic equations as $$ \frac{d\xi}{\sin\xi + \cos\xi}=\frac{d\eta}{\sin\xi}=\frac{dz}{z}. \tag{2} $$ The solution to $$ \frac{d\xi}{\sin\xi + \cos\xi}=\frac{d\eta}{\sin\xi}\quad\text{and}\quad \frac{d\xi}{\sin\xi + \cos\xi}=\frac{dz}{z} \tag{3} $$ yield, respectively, $$ \eta=\frac{1}{2}\left(\xi-\ln|\sin\xi+\cos\xi|\right)+c_1 \implies e^{2\eta-\xi}(\sin\xi+\cos\xi)=C_1, \tag{4} $$ and $$ \ln |z|-\frac{1}{\sqrt{2}}\ln\left|\tan\left[\frac{1}{2}\left(\xi+\frac{\pi}{4}\right)\right]\right|=C_2. \tag{5} $$ The general solution to the PDE $(1)$ is, therefore, $$ F(C_1,C_2)=0, \tag{6} $$ where $F$ is an arbitrary differentiable function, and $C_1$, $C_2$ are given by Eqs. $(4)$ and $(5)$.