Discussion
Let \(u(x,y) = 2f(y)\cos(x - 2y),\ (x,y) \in \mathbb{R}^{2}\), be a solution of the initial value problem
\[ \begin{aligned} 2u_{x} + u_{y} &= u \\ u(x, 0) &= \cos(x). \end{aligned} \]
Then \(f(1)\) is equal to
Answer: B
\(f(1)\) is equal to \(\frac{e}{2}\).
Substitute \(u = 2f(y)\cos(x - 2y)\). Writing \(c = \cos(x - 2y)\) and \(s = \sin(x - 2y)\):
\[ \begin{aligned} u_{x} &= -2f s, \quad u_{y} = 2f' c + 4f s \\ 2u_{x} + u_{y} &= 2f' c \end{aligned} \]
Setting this equal to \(u = 2f c\) gives \(f' = f\), so \(f(y) = Ae^{y}\). The initial condition \(u(x, 0) = 2A\cos x = \cos x\) gives \(A = \frac{1}{2}\).
So \(f(y) = \frac{1}{2}e^{y}\) and \(f(1) = \frac{e}{2}\). The value \(\frac{1}{2}\) is \(f(0)\), and \(e\) forgets the factor 2 in the given form of \(u\).
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