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

by RoRi

January 31, 2016

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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