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

by RoRi

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

5 comments

  1. Mohammad Azad says:

    It’s not needed with some trickery. b-bsin^2(ka) = bcos^2(ka) so we can cancel the cosine from the denominator. Also k should be in absolute value since nobody said it’s positive. If you don’t like this method, when solving for the constants, multiply both sides by sine or cosine to get sin^2 in one equation and cos^2 in the other and when you add them the coefficient becomes 1.

    • Anonymous says:

      Imo too, since if that is the case, then the division by cos(ka) is an invalid step.

      Also when substituting c_2 into c_1 the sign of bsin^2(ka) should be – not +.

      • Mohammad Azad says:

        It’s not needed with some trickery. b-bsin^2(ka) = bcos^2(ka) so we can cancel the cosine from the denominator. Also k should be in absolute value since nobody said it’s positive

      • Mohammad Azad says:

        It’s not needed with some trickery. b-bsin^2(ka) = bcos^2(ka) so we can cancel the cosine from the denominator. Also k should be in absolute value since nobody said it’s positive. If you don’t like this method, when solving for the constants, multiply both sides by sine or cosine to get sin^2 in one equation and cos^2 in the other and when you add them the coefficient becomes 1.

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