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24/11/21\n\nJAENEFER DOLCI; GRA 20191914\nLISTA B\n\ny''(t) + 4y(t) = g(t), \\quad y(0) = 0, \\quad y'(0) = 0\n\ng(t) = \\begin{cases}\n 1, & 0 < t < 1\\\n -1, & 1 < t < 2\\\n 0, & 2 < t\n\\end{cases}\n\n4 y(\\omega^2 + 4) + \\pi_{0} + TT_{2}\\n= + y(t - 2) + 4(y'(0) + 4y(t-1)) - \\pi_{1}\\n y(\\pi_{c}) + L^{-1}(\\omega^2 + 4) + 0\\n= L\\n\\quad 1. Y(0_o)(\\omega^2) + G(4)\\n= 1 \\Big( 1 - e^{-2a}\Big)\\n\\quad L(s, t) + 1\\n= 1 - 2e^{-2t} + 0\\n= \\qquad Y(0_p)(s^2 + 4)\\nL(\\omega^2 + 4) \\sin(s)\\n= 1-\pi_{0}i\\n\\quad L + L^{-1}(s^2 + 4)\\n= 1 - \\frac{1}{4} + L\\n= \\frac{1}{2}\\n= 2\\n\\quad L^{-1}(\\omega^2 + 4)\n 26/11/21\n\n4 TRANSFORMADA INVERSA\nL^{-1}(y(t)) = L^{-1} \\left( \\frac{1}{1 + L^{-1} \\left( \\frac{2e^{-2s}}{s^2 + 4} \\right) + L^{-1} \\left( \\frac{s^2}{s^2 + 4} \\right)} \\right)\nL^{-1} \\left( \\frac{1}{s^2 + 4} \\right) = A + B \\cos(2t)\nL^{-1} \\left( \\frac{1}{s(\\omega^2 + 4)} \\right)\\n\n1 = 1 + B + 1 = 1 + B + 1 \\frac{1}{4}\n4A = 1 \\quad A = \\frac{1}{4}\n\\n1 = A + 1 + 4A\nA = \\frac{1}{4} \\quad \\frac{1}{4}\nL^{-1} \\left( \\frac{1}{s^2} \\right) = \\frac{1}{4}\n\nL^{-1} \\left( \\frac{1}{s^2 + 1} \\right) = \\frac{1}{4}\n= 1 \\cdot 1 - 1 \\cdot \\cos(2t) = 1 - \\cos(2t)\n= \\frac{1}{4}\n\n\n4 L^2 \\quad \\quad 2 \\; \\text{TERMO (EXPANDIR)} \\quad L^{-1} \\left(\\frac{2e^{s}}{s^2}\\right)\\n= L(t - 1) \\qquad L_{2}(t-1)\\n= L^{-1} \\left( \\frac{1}{(2^2 + 4)} \\right)\n= L^{-1} \\left( \\cos(2t) \\right)\n\\quad L(t - 2) = L(t - 2)\n= L \\left( L^{-1} \\left( \\cos(2(t - 1)) \\right) \\right)\n= 2 \\cdot L(t)\n1 - \\cos(2a(t-1))u(t - 1)\n=\ - \\cdot t \\cdot u(t)\n 12/11/21\n\nL \\; 3^3 \\; \\text{TERMO EXPANDIR}\n\\frac{e^{s}}{s_{2}} + u(t - 2) = x u(t - 2)\\n\\frac{\\omega^2 + 1}{L^2+4}\\n+ L \\left( \\frac{1}{4}\\right) u_2(t-2)\\n\\frac{L^2}{X} = \nL u(t-2)\n= L \\left( 1- \\cos(2(t-1)) \\right) v_{u(t-1)}\\n1 - 1 \\cdot \\cos(2(t - 1))u(t - 1)\\n+ L u(t - 1)\\n+ L L(\\omega^2)\\n= 1- \\frac{1}{4}\n\n
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24/11/21\n\nJAENEFER DOLCI; GRA 20191914\nLISTA B\n\ny''(t) + 4y(t) = g(t), \\quad y(0) = 0, \\quad y'(0) = 0\n\ng(t) = \\begin{cases}\n 1, & 0 < t < 1\\\n -1, & 1 < t < 2\\\n 0, & 2 < t\n\\end{cases}\n\n4 y(\\omega^2 + 4) + \\pi_{0} + TT_{2}\\n= + y(t - 2) + 4(y'(0) + 4y(t-1)) - \\pi_{1}\\n y(\\pi_{c}) + L^{-1}(\\omega^2 + 4) + 0\\n= L\\n\\quad 1. Y(0_o)(\\omega^2) + G(4)\\n= 1 \\Big( 1 - e^{-2a}\Big)\\n\\quad L(s, t) + 1\\n= 1 - 2e^{-2t} + 0\\n= \\qquad Y(0_p)(s^2 + 4)\\nL(\\omega^2 + 4) \\sin(s)\\n= 1-\pi_{0}i\\n\\quad L + L^{-1}(s^2 + 4)\\n= 1 - \\frac{1}{4} + L\\n= \\frac{1}{2}\\n= 2\\n\\quad L^{-1}(\\omega^2 + 4)\n 26/11/21\n\n4 TRANSFORMADA INVERSA\nL^{-1}(y(t)) = L^{-1} \\left( \\frac{1}{1 + L^{-1} \\left( \\frac{2e^{-2s}}{s^2 + 4} \\right) + L^{-1} \\left( \\frac{s^2}{s^2 + 4} \\right)} \\right)\nL^{-1} \\left( \\frac{1}{s^2 + 4} \\right) = A + B \\cos(2t)\nL^{-1} \\left( \\frac{1}{s(\\omega^2 + 4)} \\right)\\n\n1 = 1 + B + 1 = 1 + B + 1 \\frac{1}{4}\n4A = 1 \\quad A = \\frac{1}{4}\n\\n1 = A + 1 + 4A\nA = \\frac{1}{4} \\quad \\frac{1}{4}\nL^{-1} \\left( \\frac{1}{s^2} \\right) = \\frac{1}{4}\n\nL^{-1} \\left( \\frac{1}{s^2 + 1} \\right) = \\frac{1}{4}\n= 1 \\cdot 1 - 1 \\cdot \\cos(2t) = 1 - \\cos(2t)\n= \\frac{1}{4}\n\n\n4 L^2 \\quad \\quad 2 \\; \\text{TERMO (EXPANDIR)} \\quad L^{-1} \\left(\\frac{2e^{s}}{s^2}\\right)\\n= L(t - 1) \\qquad L_{2}(t-1)\\n= L^{-1} \\left( \\frac{1}{(2^2 + 4)} \\right)\n= L^{-1} \\left( \\cos(2t) \\right)\n\\quad L(t - 2) = L(t - 2)\n= L \\left( L^{-1} \\left( \\cos(2(t - 1)) \\right) \\right)\n= 2 \\cdot L(t)\n1 - \\cos(2a(t-1))u(t - 1)\n=\ - \\cdot t \\cdot u(t)\n 12/11/21\n\nL \\; 3^3 \\; \\text{TERMO EXPANDIR}\n\\frac{e^{s}}{s_{2}} + u(t - 2) = x u(t - 2)\\n\\frac{\\omega^2 + 1}{L^2+4}\\n+ L \\left( \\frac{1}{4}\\right) u_2(t-2)\\n\\frac{L^2}{X} = \nL u(t-2)\n= L \\left( 1- \\cos(2(t-1)) \\right) v_{u(t-1)}\\n1 - 1 \\cdot \\cos(2(t - 1))u(t - 1)\\n+ L u(t - 1)\\n+ L L(\\omega^2)\\n= 1- \\frac{1}{4}\n\n