Disprove the existence of solution for the Cauchy problem $u_x-y u_y=2 u,$ subject to $u(x, 0)=f(x)$, whenever $f$ is linearly independent of $e^{2x}$

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Let me explain my attempt on the question:

Q. Consider the equation $$u_x-y u_y=2 u,$$ subject to $u(x, 0)=f(x)$. Then which of the following is/are the following are true.

(a) If the function $f$ is not a scalar multiple of $e^{2 x}$, then the solutions exist.

(b) If the function $f$ is a scalar multiple of $e^{2 x}$, then infinitely many solution exists.

(c) If the function $f$ is a scalar multiple of $e^{2 x}$, then no solutions exist.

(d) If the function $f$ is not a scalar multiple of $e^{2 x}$, then infinitely many solution exists.

Observe that the Lagrange's characteristics system $$\frac{dx}{1}=\frac{dy}{-y}=\frac{du}{2u},$$ provides a characteristic surface $$ue^{-2x}=c,$$ where $c=f(0).$ Options (b) and (c) can be dealt with the above argument.

However, I tried with functions ($f$) such as non-zero polynomials,logs, and trigonometric functions to say ''our Cauchy problem has no solution'', but I couldn't establish the non-existence even though it is evident from the unrelatable characteristics, $$x(s,t)=t+s,~y(s,t)=0,~u(s,t)=f(s)e^{-2t}.$$

I would be grateful for your correction in my arguments and any sort of help to deal options (a) and (d).