Lista 3 - Métodos Matemáticos Aplicados a Engenharia Química

12/10/21\nJAENEFER 5001 - TR000\nLISTA 3:\n1. a. -y\" + 2y' - 8y = 0; y(0) = 3; y'(0) = -12\nA = 2, b = 2, c = -8\nug. auxiliar: r^2 + 2r - 8 = 0\nr1 = -2 + √(4 - 4·1·(-8)\n2 = -2 + √32\nr2 = -2 - √32\nr1 = 2, r2 = -4\nr1 = 2\nr2 = -4\n\ny = C1 e^(r1 x) + C2 e^(r2 x)\nAPLICANDO -> y(0) = 3\n3 = C1 + C2\n3 = C1 + C2\nAPLICANDO + y'(0) = -12\ny' = C1 e^(r1·0) + C2 e^(r2·0)\n-12 = C1 r1 + C2 r2\n-12 = C1·2 + C2·(-4)\n-6 = C1 - 2C2\n3 = C1 + C2\n0 = C1 - 2C2\nC2 = 3\nC1 = 0\ny = 3 e^(2 x) 14/10/21\n(-y\" - 4y' + 4y = 0; y(-1) = 3; y'(-1) = 9)\nA = 1; b = -2; c = 3\nug. auxiliar: r^2 - 4r + 3 = 0\nr1 = +4 + √(4^2 - 4·1·3)\nr2 = +4 - √(4^2 - 4·1·3)\nr1 = +4 + 2 = 3\nr2 = 4 - 2 = 1\n2\n\n3) y = C1 e^(3x) + C2 e^(x) APLICANDO -> y(-1) = 3\n1 = C1 e^(3·(-1)) + C2 e^(-1)\n3 = C1·C2·(-1)\nAPLICANDO -> y'(-1) = 9\ny' = C1 e^(3x) + C2 e^(x)\n-3 = C1·(3e^(-3)) + C2·(e^(-1))\n-3 = C1·(-3) + C2·(-1)\n-6 = C1 + 2 C2\n0 = C1 + 2 C2\nC2 = 3\nC1 = 0\ny = 3 e^(x) 14/10/21\ny = e^(2x) + y' e^(2x)\ny' = C1 e^(x) + C2 n e^(2x) APLICANDO + y(0) = 1\nC1 = 1\n\ny = C1 e^(3x) + C2 e^(2x) APLICANDO -> y(0) = -3\n-3 = e^0 + C2 (e^0 - 0·e)\n-3 = 1 + C1(e^0)\nC1 = 0\nC2 = 2\ny = e^(x) + 2 n e^(x)\n-y\" - y' + 2y = 0; y(0) = 2; y'(0) = 1\nA = 1; b = 2; c = 0\nug. auxiliar: n^2 + 2n + 1 = 0\nr1 = -2 + √(4 - 4·1·1)\nr2 = -2 - √(4 - 4·1·1)\nr1 = -1 + √(4 - 2)\n\nr1 = -1 + √(4 - 4) = -1 + 0\ni\n-1 - x + y( 14 / 10 / 21\n\nA = 2 + C0\ny = 2 e^{c} \\cos(\\rho) + 3 e^{c} \\sen(\\rho)\nC0 = 3\n\ne - y' + 2y' = 2y' i \n y(1) = e^{r}, y'(1) = 0\n\ny'' - 2y' + 2y = 0 \\quad \\text{(apply: } r^2 - 2r + 2 = 0\n\\alpha = 1 \\quad \\beta = 2,\n\\beta_1 = 2 + \\sqrt{4 - 1.2}\n\\beta_2 = -1 - \\sqrt{4 - 1.2},\n\nn = 1 + i\\n\ny_1 = C_{1} e^{c} \\cos(\\rho) + C_{2} e^{c} \\sen(\\rho) \\quad \\text{(apply and } y_n(1) - e^{c}\n\ne^{c} = C_1 \\ \\beta = 1\n\nC0 = 1.\ne^{c} \\\\n\ny' = C_{1} \\left( e^{c} \\cos(\\rho) + C_{2} e^{c} \\sen(\\rho)\right) + C_{3} \\left( e^{c} \\sen(\\rho) + e^{c} \\cos(\\rho)\right)\n0 = C_{1} \\left( \\cos H - e^{c} \\sen(\\rho)\right) + C_{0} \\left( e'^{c} \\sen(\\rho) + \\cos(\\rho) e'^{c} \\right)\n0 = \\left( e^{c} - 1 \\right) + C_{0 \\left( -1 e^{c} \\right)}\nC{c} = 1.\ny = -e^{c} \\cos(\\rho) + e^{c} \\sen(\\rho)\ny = e^{c} \\sen(\\rho) - e^{c} \\cos(\\rho)\n 14 / 10 / 21\n\n2. y_{y}'' - y' + y(0)=0\n\ny'' + y' = 0\\text{ (apply auxiliary: } r^2 + r = 0)\n\\alpha = 0 \\to \\beta = 0 \n\\beta_1 = -1 + \\sqrt{1 - 1.0}\n\\beta_2 = -1 - \\sqrt{4 - 1.0}\n = 0\n\ny= C_{1} e^{c} + C_{2} e^{c} \\quad \\text{ (apply and } y(0) = 1)\n0 = C_1 e^0 + C_2 e^0 \\quad \n1 = C_1 + C_{2} = 1\\ \n\\quad\\ C_{_1} = 0 \n\\to C_{0} = 1 - C\\n\n0=C_{1}+C_{2}=1 \\\t\\ C_0= 1 - C_2 \\to C_{0} \\text{ (apply)}\n\\quad\\ C_1 = 1.\n\ny = 1 - e^{-1}\\n y = C_{1} e^{c} - e^{c} \\Rightarrow y = e^{c}\\n C_{1} y = e^{-c} e^{-c} e^{c} - e^{-1} \\n ( 1-e^{-1}) e^{1} \\n r^{-1} = e^{-1}\\n e^{-1} - n