Solve the following linear first order PDEs Lagrange characteristic method
$u_x + 3u_y = 5z + \tan(y-3x)$
well then from the Lagrange characteristic equations
$$\frac{dx}{1} = \frac{dy}{3} = \frac{dz}{5z + \tan(y-3x)}$$
The first part is easy from
$\frac{dx}{1} = \frac{dy}{3}$
you obtain $\psi(x,y,z) = 3x - y = c_2$
The difficult part is obtaining
$\phi(x,y,z) = ?? = c_1$
so as to obtain the general solution that is
$F(\phi(x,y,z) , \psi(x,y,z)) = 0$
any idea?
$$u_x + 3u_y +0\:u_z = 5z+\tan(y-3x)$$
Charpit-Lagrange characteristic ODEs :
$$\frac{dx}{1} = \frac{dy}{3}= \frac{dz}{0} = \frac{du}{5z+\tan(y-3x)}$$ A first obvious characteristic equation comes from $dz=0$ $$z=c_1$$ A second characteristic equation comes from solving $\frac{dx}{1} = \frac{dy}{3}$ : $$y-3x=c_2$$ A third characteristic equation comes from solving $\frac{dx}{1} = \frac{du}{5z+\tan(y-3x)}= \frac{du}{5c_1+\tan(c_2))}$ : $$u-x\:\big(5c_1+\tan(c_2)\big)=c_3$$ The general solution of the PDE on implicit form $c_3=F(c_1,c_2)$ leads to the explicit general solution : $$u(x,y,z)=x\: \big(5z+\tan(y-3x)\big)+F\big(z \:,\: y-3x \big)$$ $F$ is an arbitrary function of two variables. If some valid boundary conditions are specified, one have to determine the function $F$ according to the boundary conditions.