Find all solutions of y′′ + 25y = 0 for given initial values

Find the particular solution of the differential equation

$y'' + 25y = 0$

satisfying the initial condition $y = -1$ and $y' = 0$ when $x = 3$ .

This is a second-order linear differential equation of the form

$y'' + ay' + by = 0 \qquad \text{with} \qquad a= 0, \ \ b = 25.$

These values of $a$ and $b$ give us $d = a^2 - 4b = -100$ . So, $d < 0$ and $k = \frac{1}{2} \sqrt{-d} = 5$ . By Theorem 8.7 (pages 326-327 of Apostol) we then have

$\begin{align*} &&y &= c_1 u_1 + c_2 u_2 & u_1 = \cos (kx), \quad u_2 = \sin (kx) \\ \implies && y&= c_1 \cos(5x) + c_2 \sin (5x) \\ \implies && y'&= -5c_1 \sin (5x) + 5c_2 \cos (5x). \end{align*}$

We then use the initial conditions $y(3) = -1$ and $y'(3) = 0$ to solve for $c_1$ and $c_2$ ,

$\begin{align*} 0 = y'(3) &= -5c_1 \sin 15 + 5c_2 \cos 15 & \implies && c_1 = c_2 \cot 15 \\ -1 = y(3) &= c_1 \cos 15 + c_2 \sin 15 & \implies && c_2 = -\sin 15, \ \ c_1 = -\cos 15. \end{align*}$

Therefore,

$y = -\cos 15 \cos (5x) - \sin 15 \sin (5x).$

Stumbling Robot

Find all solutions of y′′ + 25y = 0 for given initial values

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