Lista 4 - Métodos Matemáticos Aplicados a Engenharia Química

JAENEFER 201, 1000\nlista 4 :\n1. y\" - y = 8 e^2 t + 2 e^t\n\nDe acordo com os princípios da superposição, podemos resolver o problema, um de dois partes:\n(i) y\" - y = 8 e^t - p(e)\n(ii) y\" - y = 2 e^t - p(e)\ny\" - 11 = 8 e^t\nC_m m = 8 e^t\n\nCom considerando razoavelmente valores particulares:\n(i) y_p1 = A e^t + B e^t\n\nyp1 = A e^2 t + B e^t + C e^t\nyp1\" = 8 e^2 t\nyp1 = (4 A) + (2 A + 2 B) = 8\n{4A = 8}\n-2A + 2B = 0\n\nA = 2\nB = -1/12\nB = -2\n\ny_p = 2A e^t - 2t e^2 t 20 10 21\nφ e^x\n2 e^t c + A e^t \nypa = A e^T\n\nyp1 = A e^2 x + e^t + C e^t\nyp1 = A e^t + 2 A e^t\nyp1 + y_p = e^t - 2 A e^t - 2 B e^{t}\n= e^{t} (A e^x - B e^{2 x})\n\nEx 3:\n2. y\" - 3y' = -2y + e^{sin x}\na = 1, b = 3, c = 2\nup. aliminar: r^2 - 3r - 2 = 0\n\nr1 = 3 + √1\np x = 3 - 1√\n\ny = 3 + (β - 4) + 2 p DIXANDO C§ES EM EVIDÊNCIA:\n(-2B - 3A + 3B + 2A^2) e^{se^t} + (2A - 3A - 3B + B) e^5 e^t = 0 \n\n1 + B2 = 1\n-A + B = 0\n\nA = -1, B = 1\n\ny_p = 1 cos x e^{2t} - 1 sin x e^{2t}\n\ny_p = B1(cos x - sin x)\n\ny = C_1 e^{x} + C_2 e^{t} + (cos x - sin x) e^{2t}\n3. y\" + y = 2 e^{t} e^{t} y(0) = 0 y(0) = 1\n y = C1 cos(αx) + C2 sen(αx)\n\ny'' + p y' + q y = 5 sen(x) + 5 cos(x)\n\ny(0) = 1 y'(0) = 2\nA = 2, C1 = 2\nC2 = -2\n\nr1 = -2 - √4 = -4\nr2 = -2 + √4 = 0\n\nm1 = -1 - i m2 = -1 + i\n\nyp = C1 e^(-x) cos(αx) + C2 e^(-x) sen(αx)\n\nyp' = ...\n\nyp'' = ...\n\n(-A sen(x) - B cos(x)) + 2(A cos(x) - B sen(x)) + 2(A sen(x) + B cos(x)) = 5 sen(x) + 5 cos(x)\n\nA - 2B = 5\n\nA + B = 5\n\n5A = 15\n\nA = 3\n\ny(0) = 1 → 1 = C1 - 1\nC1 = 2 y'' - y' + 2y'' = 5 sen(x) + 5 cos(x)\n\ny(0) = 1 y'(0) = 2\nA = -2, C1 = C2\n\nyp = C1 e^(-x) cos(αx) + C2 e^(-x) sen(αx)\n\ny' = A cos(αx) + B sen(αx)\n\nyp'' = -A sen(x) - B cos(x)\n\n(-A sen(x) - B cos(x) - (A sen(x) + B cos(x)) = sen(x)\n\n-2A = 1 → A = -1 B = 0\n\nypi = -1 sen(x) y'' - y' - 5sen(x) + e^(-x)\n\ny(0) = 2\nC = C1 + C2 + 3\nC0 = 1\n\ny = 2e^(-x) cos(αx) + e^(-x) sen(αx) + 3sen(x) - cos(x)\n\ny' = A sen(x) + B cos(x)\n\nyp = -A sen(x/2) - B sen(x/2)\n-2A = 1 → A = -1 B = 2 (ii) y_p = A_e^{x} \n y_p'' = 2A_e^{x} \n y_p''' = 4A_e^{x} \n 3A_e^{x} - e^{-x} \n \n 4Ae^{x} - e^{-x} \n y_p'' = -1 e^{x} \n 3A_e^{x} - 1 \n A = -1 \n 3 \n \n y_p + y_p' = -1 e^{-x} \sin x + C_1e^{x} + C_2e^{-x} \n 3 2 \n (y_p' + y_p'') = -2 e^{x} - 1 \cos x + C_1e^{x} + C_2e^{x^2} \n 3 2 \n y(0) = 1 \n -1 + C_1 + C_2 = 1 \n 3 \n C_1 + C_2 = 4 \n 3 \n C_1 - C_2 = 1 \n 6 \n \int (C_1 + C_2 = 4/6) 3 + C_3 = 6 \n C_1 - C_2 = 1/6 \n \n C_1 = 3/12 \n C_2 = 7 \n C_1 = 3/4 \n \n y_p = e^{-x} \n 12 \n 12