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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*}

Thus, we have a unique solution y = c_1 \cos (kx) + c_2 \sin (kx) with c_1 and c_2 as given above. \qquad \blacksquare

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