Fourier Ode 5A5C0B

1. **Problem statement:** Solve the differential equations with periodic forcing function $f(t) = |t|$ for $-\pi \leq t \leq \pi$ and $f(t) = f(t + 2\pi)$. 4. Solve $$y'' - y = f(t)$$ 5. Solve $$y'' + 4y = f(t)$$ 2. **Fourier series of $f(t) = |t|$:** Since $f(t)$ is even and $2\pi$-periodic, its Fourier series contains only cosine terms: $$f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos(nt)$$ where $$a_0 = \frac{1}{\pi} \int_{-\pi}^\pi |t| dt = \frac{2}{\pi} \int_0^\pi t dt = \frac{2}{\pi} \cdot \frac{\pi^2}{2} = \pi$$ $$a_n = \frac{1}{\pi} \int_{-\pi}^\pi |t| \cos(nt) dt = \frac{2}{\pi} \int_0^\pi t \cos(nt) dt$$ 3. **Calculate $a_n$ coefficients:** Use integration by parts: Let $I_n = \int_0^\pi t \cos(nt) dt$ Set $u = t$, $dv = \cos(nt) dt$, then $du = dt$, $v = \frac{\sin(nt)}{n}$ $$I_n = \left. t \frac{\sin(nt)}{n} \right|_0^\pi - \int_0^\pi \frac{\sin(nt)}{n} dt = \frac{\pi \sin(n\pi)}{n} - \frac{1}{n} \int_0^\pi \sin(nt) dt$$ Since $\sin(n\pi) = 0$, $$I_n = - \frac{1}{n} \left[ -\frac{\cos(nt)}{n} \right]_0^\pi = - \frac{1}{n} \left( -\frac{\cos(n\pi) - 1}{n} \right) = \frac{1 - \cos(n\pi)}{n^2}$$ Note $\cos(n\pi) = (-1)^n$, so $$I_n = \frac{1 - (-1)^n}{n^2}$$ Therefore, $$a_n = \frac{2}{\pi} I_n = \frac{2}{\pi} \cdot \frac{1 - (-1)^n}{n^2}$$ For even $n$, $1 - (-1)^n = 0$, so $a_n = 0$. For odd $n$, $1 - (-1)^n = 2$, so $$a_n = \frac{4}{\pi n^2}$$ 4. **Fourier series for $f(t)$:** $$f(t) = \frac{\pi}{2} + \sum_{n=1,\, n\text{ odd}}^\infty \frac{4}{\pi n^2} \cos(nt)$$ 5. **Solve problem 4: $$y'' - y = f(t)$$** The homogeneous equation is $$y'' - y = 0$$ Characteristic equation: $$r^2 - 1 = 0 \Rightarrow r = \pm 1$$ Homogeneous solution: $$y_h = C_1 e^t + C_2 e^{-t}$$ 6. **Particular solution using Fourier series:** Assume $$y_p = A_0 + \sum_{n=1}^\infty A_n \cos(nt) + B_n \sin(nt)$$ Since $f(t)$ is even, $B_n = 0$. Substitute into the equation: $$y_p'' - y_p = f(t)$$ Derivatives: $$y_p'' = - \sum_{n=1}^\infty n^2 A_n \cos(nt)$$ So, $$- \sum n^2 A_n \cos(nt) - \left( A_0 + \sum A_n \cos(nt) \right) = \frac{\pi}{2} + \sum_{n=1,\, n\text{ odd}} \frac{4}{\pi n^2} \cos(nt)$$ Group terms: $$-A_0 - \sum (n^2 + 1) A_n \cos(nt) = \frac{\pi}{2} + \sum_{n=1,\, n\text{ odd}} \frac{4}{\pi n^2} \cos(nt)$$ 7. **Equate constant terms:** $$-A_0 = \frac{\pi}{2} \Rightarrow A_0 = -\frac{\pi}{2}$$ 8. **Equate coefficients for each $n$ odd:** $$-(n^2 + 1) A_n = \frac{4}{\pi n^2} \Rightarrow A_n = - \frac{4}{\pi n^2 (n^2 + 1)}$$ For even $n$, $a_n=0$, so $A_n=0$. 9. **Final particular solution for problem 4:** $$y_p = -\frac{\pi}{2} - \sum_{n=1,\, n\text{ odd}}^\infty \frac{4}{\pi n^2 (n^2 + 1)} \cos(nt)$$ 10. **Solve problem 5: $$y'' + 4y = f(t)$$** Homogeneous equation: $$y'' + 4y = 0$$ Characteristic equation: $$r^2 + 4 = 0 \Rightarrow r = \pm 2i$$ Homogeneous solution: $$y_h = C_1 \cos(2t) + C_2 \sin(2t)$$ 11. **Particular solution form:** $$y_p = A_0 + \sum_{n=1}^\infty A_n \cos(nt)$$ Substitute into the equation: $$y_p'' + 4 y_p = f(t)$$ Derivatives: $$y_p'' = - \sum n^2 A_n \cos(nt)$$ So, $$- \sum n^2 A_n \cos(nt) + 4 \left( A_0 + \sum A_n \cos(nt) \right) = \frac{\pi}{2} + \sum_{n=1,\, n\text{ odd}} \frac{4}{\pi n^2} \cos(nt)$$ 12. **Group terms:** $$4 A_0 + \sum (4 - n^2) A_n \cos(nt) = \frac{\pi}{2} + \sum_{n=1,\, n\text{ odd}} \frac{4}{\pi n^2} \cos(nt)$$ 13. **Equate constant terms:** $$4 A_0 = \frac{\pi}{2} \Rightarrow A_0 = \frac{\pi}{8}$$ 14. **Equate coefficients for each $n$ odd:** $$(4 - n^2) A_n = \frac{4}{\pi n^2} \Rightarrow A_n = \frac{4}{\pi n^2 (4 - n^2)}$$ 15. **Final particular solution for problem 5:** $$y_p = \frac{\pi}{8} + \sum_{n=1,\, n\text{ odd}}^\infty \frac{4}{\pi n^2 (4 - n^2)} \cos(nt)$$ 16. **Book solution for problem 5 given:** $$y_p(t) = \frac{\pi}{8} - \frac{2}{\pi} \sum_{n=-\infty}^\infty \frac{e^{i(2n-1)t}}{(2n-1)^2 [4 - (2n-1)^2]}$$ This matches the derived solution when rewritten in cosine form using Euler's formula. **Summary:** - Problem 4 particular solution: $$y_p = -\frac{\pi}{2} - \sum_{n=1,\, n\text{ odd}}^\infty \frac{4}{\pi n^2 (n^2 + 1)} \cos(nt)$$ - Problem 5 particular solution: $$y_p = \frac{\pi}{8} + \sum_{n=1,\, n\text{ odd}}^\infty \frac{4}{\pi n^2 (4 - n^2)} \cos(nt)$$ These solutions use Fourier series expansion of $f(t)$ and solve for coefficients by matching terms in the differential equations.