Using the method of characteristics with an extra term

Question

How do you apply the method of characteristics to get the solution to the following PDE:

$$xU_x-yU_y-xU=x^2y$$

Answer 1

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dx}{dt}=x$ , letting $x(0)=1$ , we have $x=e^t$

$\dfrac{dy}{dt}=-y$ , letting $y(0)=y_0$ , we have $y=y_0e^{-t}=\dfrac{y_0}{x}$

$\dfrac{dU}{dt}=xU+x^2y=e^tU+y_0e^t$ , we have $U(x,y)=f(y_0)e^{e^t}-y_0=f(xy)e^x-xy=F(xy)e^x$

Answer 2

$\begin{cases} \dfrac{dx}{dt}=x\\ \dfrac{dy}{dt}=-y\\ \dfrac{dU}{dt}=xU+x^2y \end{cases}$ is the same as : $\quad \frac{dx}{x}=\frac{dy}{-y}=\frac{dU}{xU+x^2y}=dt $

A first characteristic equation comes from $\frac{dx}{x}=\frac{dy}{-y}$ :

$$xy=c_1$$

A second characteristic equation comes from $\frac{dx}{1}=\frac{dU}{U+xy}=\frac{dU}{U+c_1}$ :

$$e^{-x}(U+c_1)=c_2$$ The general solution of the PDE expressed on inplicit form $c_2=F(c_1)$ is : $$e^{-x}(U+xy)=F(xy)$$ $F$ is an arbitrary function, to be determined in order to satisfy some boundary condition (not specified in the wording of the question). $$\boxed{U(x,y)=e^xF(xy)-xy}$$ This canot be simplified until the boundary condition be given.