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Find constants to make solutions of y′′ + k2y = 0 have specific properties

There exists a positive real number k such that

    \[ y'' + k^2y = 0 \]

has solutions y = f(x) with f(0) = f(3) = 0 and f(x) < 0 for all x in the open interval (0,3). Compute the value of k and find all solutions of the equation.


Since f(x) is a solution of y'' + k^2 y = 0 we know

    \[ f(x) = A \cos (kx) + B \sin (kx) \]

for some constants A and B. We are given f(0) = 0,

    \[ f(0) = 0 \quad \implies \quad A \cos 0 = 0 \quad \implies \quad A = 0. \]

Therefore,

    \[ f(x) = B \sin (kx),\]

with B \neq 0 (since if B = 0 then f(x) = 0 for all x contradicting that f(x) < 0 on 0 < x < 3). Therefore,

    \[ B \sin (3k) = 0 \quad \implies \quad \sin (3k) = 0 \quad \implies \quad k = \frac{n \pi}{3} \]

for n \in \mathbb{Z}_{\geq 0}. (We know n > 0 since k > 0.)
Then, since f(x) < 0 on 0 < x < 3, we know n = 1, otherwise f would change sign on the interval. Hence,

    \[ k = \frac{\pi}{3}, \]

and the solutions are

    \[ y = A \sin \frac{\pi x}{3}, \qquad A < 0. \]

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