1. Use Fourier series to solve the differential equation
\[
\frac{d^{2} y}{d t^{2}}+25 y=f(t)
\]
where \( f(t)=|t|,-\pi \leq t \leq \pi, f(t+2 \pi)=f(t) \).

Answer

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Answer

\(y(t) = \sum_{n=1}^\infty \left[\frac{-b_n}{25 - n^2} \cos(\omega_n t) - \frac{a_n}{25 - n^2} \sin(\omega_n t)\right]\)

Steps

Step 1 ：\(f(t) = \frac{a_{0}}{2} + \sum_{n=1}^\infty \left[a_n \cos(\omega_n t) + b_n \sin(\omega_n t)\right], \quad \omega_n = n\)

Step 2 ：\(a_0 = \frac{2}{2\pi}\int_{-\pi}^{\pi} |t| dt, \) \(a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} |t| \cos(nt) dt, \) \(b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} |t| \sin(nt) dt\)

Step 3 ：\(y(t) = \sum_{n=1}^\infty \left[\frac{-b_n}{25 - n^2} \cos(\omega_n t) - \frac{a_n}{25 - n^2} \sin(\omega_n t)\right]\)

1. Use Fourier series to solve the differential equation\[\frac{d^{2} y}{d t^{2}}+25 y=f(t)\]where \( f(t)=|t|,-\pi \leq t \leq \pi, f(t+2 \pi)=f(t) \).

Answer

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1. Use Fourier series to solve the differential equation
\[
\frac{d^{2} y}{d t^{2}}+25 y=f(t)
\]
where \( f(t)=|t|,-\pi \leq t \leq \pi, f(t+2 \pi)=f(t) \).