Consider the following PDE: $$-3U_x+4U_y=0$$ where $U=3x$ on curve $y=x+1$.
First we invoke the method of characteristics: $$\frac{dx}{-3}=\frac{dy}{4}=\frac{dU}{0}.$$ From this we get $c_1=\frac{x}{3}+\frac{y}{4}$ and $U(x,y)=c_2$. So $F(c_1)=c_2$ or $F(\frac{x}{3}+\frac{y}{4})=u(x,y)$.
Applying the condition, we know that $3x=u(x,x+1)=F(\frac{x}{3}+\frac{x+1}{4})=F(\frac{x+3}{12}).$ So $F(\frac{x+3}{12})=3x.$ This is now where I'm stuck.
The general solution of $-3U_x+4U_y=0$
obviously is $U(x,y)=F(\frac{x}{3}+\frac{y}{4}) \space$ where $F$ is any derivable function.
I don't understand what you mean "incorporate the curve". If you are talking of a boundary condition on the curve $y=x+1$, please, make clear what is the boundary condition.
With the condition $U(x,y=x+1)=3x\space$ one have to find a function $F\space$ so that $$F(\frac{x}{3}+\frac{x+1}{4})=3x$$ $$F(\frac{7x}{12}+\frac{1}{4})=3x$$ It is easy to see that the equality is obtained witth a function $F(X)=aX+b$ where $a=\frac{36}{7}$ and $b=-\frac{36}{7} \frac{1}{4}$ because : $$\frac{36}{7}(\frac{7x}{12}+\frac{1}{4})-\frac{9}{7}=3x$$ As a consequence, the final result is : $$U(x,y)=\frac{36}{7}(\frac{x}{3}+\frac{y}{4})-\frac{9}{7}$$