Find constants so that a differential equation has nontrivial solutions

by RoRi

January 31, 2016

Consider the second-order differential equation

$y'' + ky = 0.$

Find all values for the constant $k$ such that the equation has a nontrivial solution $y = f_k (x)$ such that

$f_k (0) = f_k (1) = 0.$

For each such $k$ , determine the corresponding solution $y = f_k(x)$ .

If $k = 0$ , then

$y'' = 0 \quad \implies \quad y = c_1 x + c_2.$

The condition $f(0) = f(1) = 0$ then implies $c_1 = c_2 = 0$ , so only the trivial solution is possible.

If $k > 0$ , then then $y'' + ky = 0$ is of the form

$y'' + ay' + by = 0 \qquad \text{with} \qquad a = 0, \ \ b = k.$

Therefore, $d = a^2 - 4b = -4k$ and $d < 0$ since $k> 0$ . Hence the solutions are of the form

$y = c_1 \cos ( \sqrt{k} x) + c_2 \sin (\sqrt{k} x).$

Then, $f(0) = 0$ implies $c_1 = 0$ , and $f(1) = 0$ implies

$c_2 \sin (\sqrt{k} x) = 0 \quad \implies \quad \sqrt{k} = n \pi \quad \implies k = n^2 \pi^2,$

where $n$ is an integer, or $c_2 = 0$ . But, if $c_2 =0$ then we have the trivial solution.

If $k < 0$ , then $y'' + ky = 0$ is of the form

$y'' + ay' + by = 0 \qquad \text{with} \qquad a = 0, \ \ b = k.$

Therefore, $d = a^2 - 4b = -4k$ and $d > 0$ since $k < 0$ . Hence, the solutions are of the form

$y = c_1 e^{\sqrt{-k} x} + c_2 e^{-\sqrt{-k} x}.$

The condition $f(0) = 0$ implies $c_1 = -c_2$ and the condition $f(1) = 0$ implies

$c_1 e^{\sqrt{-k}x} - c_1 e^{-\sqrt{-k}x} = 0 \quad \implies \quad c_1 = 0 \quad \text{or} \quad e^{2\sqrt{-k} x} - 1 = 0.$

But this latter solution is only possible if $k = 0$ , contradicting that $k < 0$ . Hence, only the trivial solution is possible for $k < 0$ .

Putting this all together, the only nontrivial solution is for $k > 0$ , and is given by

$y = c \sin (n \pi x), \qquad n \in \mathbb{Z}.$

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