I have the partial differentiation question $$du/dx -3xdu/dy = 0$$ and initial conditions $$u(x,0)=cos2x$$ I have $$A=y+3x$$ and $$u(x,y)=B$$ from integration.
So I said $$u(x,y)=F(y+3x)$$ thus $$u(x,0)=F(3x)=cos2x$$
How would I find out what F is too sub back in or how would I do the solution? and I also would the characteristic curve be $y=y(-3x,C)$ but I have a feeling this is wrong. Thank you
Supposing that there is a typo in $\quad\frac{\partial u}{\partial x}-3x\frac{\partial u}{\partial y}=0\quad$ and that the correct PDE is : $$\frac{\partial u}{\partial x}-3\frac{\partial u}{\partial y}=0$$ If not, your calculus for the characteristic $\quad A=y+3x\quad$ would be false.
With the corrected equation, I agree that the general solution is : $$u(x,y)=F(y+3x)$$
CONDITION :
$$u(x,0)=F(3x)=\cos(2x)$$ Let $\quad X=3x\quad\implies\quad F(X)=\cos(\frac23 X)$
So, the function $F(X)$ is determined. We put it into the above general solution where $X=y+3x$ : $$u(x,y)=\cos\left(\frac23 (y+3x)\right)$$