Why did I get 1/0 in the equation of characteristics?

• 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 $\frac{dy}{dx}=\frac{B\pm\sqrt{B^2-4AC}}{2A}=\frac{1}{0}$

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.

• :( ${}{}{}{}{}$ – Isabella Apr 3 '18 at 20:17
• I cannot understand your attempt. – JJacquelin Apr 7 '18 at 10:34
• @JJacquelin My professor told us that computing $\frac{dy}{dx}=\frac{B\pm\sqrt{B^2-4AC}}{2A}$ give us the equations of its characteristics. For this particular case, $A=0=C$ (which are the coefficients of $u_{xx}$ and $u_{yy}$ respectively) and $B=1$ (coefficient of $u_{xy}$), – Isabella Apr 7 '18 at 20:25

$$u_{xy}+u_x+u_y=3x$$ 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 : $\quad v_\lambda(x,y)=e^{\lambda x-\frac{\lambda}{1+\lambda}y}$
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.