Consider problem: $U_x+3U_y=xy, U(0,y)=0$, I proceed as follow:
$$\frac{dx}{1}=\frac{dy}{3}=\frac{du}{xy}$$
Integrate on first two terms:
$$x-\frac{1}{3}y=C \qquad \color{red}{(*)}$$
Integrate on $\frac{du}{xy} = \frac{dx}{1}$: Error! You must remember that in the equation of characteristics $x$ and $y$ are not independent but connected by (*).
$$U =x^3-\frac{C}{2}x^2+D = \frac{3}{2}x^3+\frac{1}{2}yx^2+D $$
D must be constant along the integral curve hence $D=\phi(x-1/3y)$
Hence, the general solution is $U =\frac{3}{2}x^3+\frac{1}{2}yx^2+\phi(x-1/3y)$.
Impose initial condition: $U(0,y)=0$ we have $\phi(-1/3y)=0$.
Update:
Hi professor, I have fixed the integration part, but I still have question about the constant D when integrating on U. Should I include this $D=\phi(x-\frac{1}{3}y)$ in the solution to IVBP?
