$\frac{\partial u}{\partial x} + 2 \frac{ \partial u}{\partial y} = 1 + u $
Where $ u = \sin x$ on $ y = 3x+1 $
Working out so far
$\frac{dx}{1} = \frac{dy}{2} = \frac{du}{1+u}$
Integrating dx and dy gives $c_1 = 2x - y $
I'm not sure how to calculate $c_2$? Do you integrate dx and du or dy and du?
$$\frac{\partial u}{\partial x} + 2 \frac{ \partial u}{\partial y} = 1 + u $$ On the characteristic curves : $$\frac{dx}{1}=\frac{dy}{2}=\frac{du}{u+1}$$ A first characteristic equation comes from $\frac{dx}{1}=\frac{dy}{2}$ : $$2x-y=c_1$$ A second characteristic equation comes from $\frac{dx}{1}=\frac{du}{u+1}$ : $$(u+1)e^{-x}=c_2$$ The general solution of the PDE expressed on implicit form $c_2=F(c_1)$ is : $$(u+1)e^{-x}=F(2x-y)$$ $F$ is an arbitrary function. $$u(x,y)=-1+e^xF(2x-y)$$ The function $F$ has to be determined according to the condition $u(x,3x+1)=\sin(x)$
$\sin(x)=-1+e^xF(2x-(3x+1))=1+e^xF(-x-1)$
Let $X=-x-1\quad;\quad x=-X-1$
$\sin(-X-1)=-1+e^{-X-1}F(X)$ $$F(X)=\big(1-\sin(X+1)\big)e^{X+1}$$ Now the function $F(X)$ is determined. We put it into the above general solution where $X=2x-y$ :
$$u(x,y)=-1+e^x\big(1-\sin(2x-y+1)\big)e^{2x-y+1}$$ $$u(x,y)=-1+\big(1-\sin(2x-y+1)\big)e^{3x-y+1}$$