Avaliação 1 - Recuperação - 2023-2

Loco: a_{2n} = \frac{a_0}{(2n)!} \quad a_{2n-1} = \frac{b_0}{(2n-1)!}/\!/ \int \omega: y(\lambda) = x^{-1/2}a_0 \left(1 + \sum_{n=1}^{\infty} \frac{x^{2n}}{(2n)!}\right) + x^{-1/2}b_0 \left( \lambda + 1 \sum_{n=1}^{\infty} \frac{x^{2n-1}}{(2n-1)!}\right) 8) y''+ 2y' = \sin T: U_n + 1: (U_0 - U_n) 2) y'' { + 4\omega} y = \frac{1}{2} \sin T: U_n { + y}\omega { * \omega} U_n s^5E - 1 + 4[E = \frac{-e^{-n5}}{s^5+1} + 1 + \frac{e^{n5}}{s^{5+4}}] F = \frac{2}{s^{4+1}} + \frac{e^{-n5}}{s^{5+4}} + \frac{e^{-n5}}{s^{2+1}} \cdot \frac{1}{s^{5+4}} = F = \frac{2}{s^{1+5}} + \frac{e^{-n5}}{2} + \frac{e^{-n5}}{2} \cdot \frac{1}{s^{4-1}} \cdot \frac{2}{s^{5+4}} \int_0^r \sin T \cdot \sin T = \int_0^r \sin T \cdot \sin (T-T)| d\int \Psi { - 2\sin T}(\cos T - 1) 100: y(t) + \sin 2T + U_1\cdot \sin (2(T-T1) + U_1{ -2 \sin (T-\pi) [U\cos (T-\pi) - 1] 2 + U_1 \frac{\sin T}{2} - \frac{U_1 (m)}{3, } +\infty} 9) f(x) = \left\{ \begin{array}{ll} x^{-3} \quad \text{se} \quad -1 \le x \le 0 \\ 1 \quad \text{se} \quad 0 \le x < 1 \end{array} \right. a_0 = \int_{-1}^1 f(x)dx = \int_{-1}^0 (x+1)dx - \int_{-1}^0 dx = \int_{-1}^0 dx + \int_{-1}^0 xdx = \frac{2-1}{2} = \frac{3}{2} a_n = \int_{-1}^1 f(x)\cos n\pi xdx = \int_{-1}^0 (1+x)\cos n\pi xdx + \int_{-1}^0 \cos(n\pi x)dx = \int_{-1}^0 \cos n\pi x dx - 1\int_0^{\infty} \cos(n\pi x) xdx = \frac{ 2\sin n\pi}{n} + \frac{\cos(n\pi)}{n^2} + \frac{x \sin(n\pi)}{n^3} = \frac{\cos(n\pi) - 1}{n^2}\\ \forall \quad n \in \mathbb{N} B - \frac{1}{n} \quad \forall \quad n \, \text{pon} \quad a_n = 0 b_n = -\int_{-1}^1 f(x)\sin(n\pi x)dx = \frac{-\cos(n\pi)}{n} \bigg|_{-1}^0 + \frac{-x\cos(n\pi)}{n^2} \bigg|_{0}^{-1} + \frac{\sin(n\pi)}{n^3} \bigg|_{-1}^{0} = \frac{-1}{n}\int(0) - \cos(n\pi x) + \sin(n\pi x)n\cancel{\ni}\\ b_n = \frac{(-1)^{n+1}}{\pi n} \text{Série de Fourier:} {f(\lambda) \sim \frac{3}{2}} + \sum_{n=1}^{\infty} \frac{-2 \cdot x^{2 n} \cos(2\pi-1n\pi x)}{n^2*\pi} + {(-1)^n \frac{-8 m (n\pi)}{n^2}} 10) U_{xx} + Cu_{yy} = 0 seja \quad U(x,y) = X(x)\cdot Y(\gamma) U_{xx} = X''Y; \quad U_{yy} = X Y'' 100: \quad X''Y + XY'' = 0 + Y''= \frac{-Y}{5} = \lambda \frac{X''}{X} = 0 \text{, com} \quad X(0) \text{e} \quad X(1) = 0 \text{Caso} \lambda = 0 X'' = 0 \implies X(\gamma) = Ax + b \cancel{X(1):0} = 0 \rightarrow b = 0\en \rightarrow \cancel{X(n)=0} \rightarrow 0 \rightarrow b=0 \rightarrow \text{se} \rightarrow\cancel{\text{trivial}} \text{Caso} \lambda > 0 X(\Pi): = C_1 e^{\lambda/\pi x} + C_2 e^{-\lambda}\pi x X(0) = 0 \rightarrow 0: C_1 + C_2 \no \cancel{\rightarrow} X(\gamma) = 0 \rightarrow 0 \rightarrow C_1 \lambda - C_2 \cancel{\lambda} \rightarrow \lambda : C_1 \ = \ C_2 := 0\text{Com}\ \text{onde}\ \cancel{C_1 = C_2: 0} \text{Caso} \lambda < 0 \lambda(x): = a{\lambda}\cos [ ~ \lambda x] + b_n\sin[\lambda^{-2} \pi x] X( \cancel{\int_0^1 }) - \lambda = x \cancel{x (\gamma) = \sin n \lambda = 0 \rightarrow \sin -1 = 0 \rightarrow \,\lambda = -n^2 \pi^2} \where \quad Y_n: \sin(n\pi x) Y'' = -\lambda Y = 0 \rightarrow \cancel{Y'' = -n^2 \pi^2 Y = 0 \text{com} \quad Y(0)\cancel{=0}} \text{trivial} y(y) = c_1 e^{n\pi y} + c_2 e^{-n\pi y} y(0) = 0 : 0 = c_1 + c_2 0 c_2 = -c_1 logo: y(y) = 2c_1 \sinh(n\pi y) = A_n \sinh(n\pi y) Como u(x,y) = X(x) Y(y) => U(x,y) = A_n \sin(n\pi x) \sinh(n\pi y) U(x,y) = x-x^3/3 - A_n\sin(n\pi x)\sinh(n\pi y) (Série de Fourier) A_n\sinh(n\pi) = 2 \int_0^1 (x-x^3)\sin(n\pi x)dx A_n\sinh(n\pi) = 2\frac{\pi \sinh(n\pi) - 2\csch(n\pi)+2}{n^2\pi^3} se n par \to \cos(C_n) = 1 \to A_n = 0 se n ímpar \to A_n = \frac{8}{\sinh(n\pi)n^3\pi^3} logo: U(x,y) = 0 para n par U(x,y) = \frac{8 \sin(n\pi x) \sinh(n\pi y)}{\sinh(n\pi)n^3\pi^3}