Why did I get $\frac {1\pm1}0$ in the equation of characteristics of $u_{xy}+u_x+u_y=3x$? 
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*Consider the partial differential equation $u_{xy}+u_x+u_y=3x$. Obtain the equation of its characteristics and reduce to the canonical form.
Attempt
The equation of its characteristics is given by $\frac{dy}{dx}=\frac{B\pm\sqrt{B^2-4AC}}{2A}$
For this particular case, $=0=$ (which are the coefficients of $_{}$ and $_{}$ respectively) and $=1$ (coefficient of $_{}$) so
the equation of its characteristics yields to $\frac{dy}{dx}=\frac{B\pm\sqrt{B^2-4AC}}{2A}=\frac{1\pm1}{0}$ 
What can I do here? This was not expected.. I can't go further, the next step was integrate respect to $y$.  
Any kind of help is greatly appreciated.
 A: $$u_{xy}+u_x+u_y=3x$$
Let $\quad u(x,y)=\frac32 x^2+v(x,y)\quad$ which leads to :
$$v_{xy}+v_x+v_y=0$$
Search for particular solutions on the form $u=X(x)Y(y)$
$X'Y'+X'Y+XY'=0\quad;\quad\frac{X'}{X}=-\frac{Y'}{Y+Y'}=\lambda \qquad\begin{cases}
X=e^{\lambda x} \\
Y=e^{-\frac{\lambda}{1+\lambda}y}\end{cases}$
Particular solutions : $\quad v_\lambda(x,y)=e^{\lambda x-\frac{\lambda}{1+\lambda}y}$
The general solution is any linear combination of the particular solutions. 
General solution expressed on the form of integral :
$$v(x,y)=\int f(\lambda)e^{\lambda x-\frac{\lambda}{1+\lambda}y}d\lambda$$
$f(\lambda)$ is an arbitrary function, in so far the integral be convergent.
$$u(x,y)=\frac32 x^2+\int f(\lambda)e^{\lambda x-\frac{\lambda}{1+\lambda}y}d\lambda$$
Or, equivalently :
$$u(x,y)=\frac32 x^2+\int g(\mu)e^{-\frac{\mu}{1+\mu}x+\mu y}d\mu$$
The function $f(\lambda)$ , or $g(\mu)$ , as well as the bounds of the integral, have to be determined according to the boundary conditions. This generally draw to solve an integral equation. Without well defined boundary conditions no further calculus is possible, even not to say if it is possible to analytically solve it and if there is a closed form for the solution.
