Prove there is exactly one solution of y′′ + k2y = 0 satisfying given constraints

Consider the second-order differential equation

$y'' + k^2 y = 0.$

If $(a,b)$ is a point in the plane and if $m \in \mathbb{R}$ is a given real number, prove that this differential equation has exactly one solution whose graph passes through the point $(a,b)$ and has slope $m$ . Consider also the case $k = 0$ .

Proof. If $k = 0$ , then $y'' = 0$ implies $y = c_1 x +c_2$ , and $y(a) = b$ implies $c_1 a + c_2 = b$ . This gives us $c_2 = b - c_1 a$ . Then, $y' = c_1 = m$ so

$y = mx + b - c_1a = m(x-a) + b.$

If $k \neq 0$ , then $k^2 > 0$ so $a^2 - 4b = -4k^2 < 0$ . This means the solutions are of the form

$\begin{align*} && y &= c_1 \cos (kx) + c_2 \sin (kx) \\ \implies && y' &= k c_2 \cos (kx) - k c_1 \sin (kx). \end{align*}$

The condition $y(a) = b$ then gives us

$\begin{align*} y(a) = b && \implies && b &= c_1 \cos (ka) + c_2 \sin (ka) \\ && \implies && c_1 &= \frac{b - c_2 \sin (ka)}{\cos (ka)}. \end{align*}$

The condition $y'(a) = m$ give us

$\begin{align*} && y'(a) &= b \\ \implies && kc_2 \cos (ka) - kc_1 \sin (ka) &= m \\ \implies && kc_2 \cos (ka) - k \left( \frac{b - c_2 \sin (ka)}{\cos (ka)} \right) \sin (ka) &= m \\ \implies && \frac{c_2 \cos^2 (ka) - b \sin (ka) + c_2 \sin^2 (ka)}{\cos (ka)} &= \frac{m}{k} \\ \implies && \frac{c_2 - b \sin (ka)}{\cos (ka)} &= \frac{m}{k} \\ \implies && c_2 &= \left( \frac{m}{k} \right) \cos (ka) + b \sin (ka) \\ \implies && c_1 &= \frac{b - \frac{m}{k} \cos (ka) \sin (ka) + b \sin^2 (ka)}{\cos (ka)}. \end{align*}$