Lista 7 - Métodos Matemáticos Aplicados a Engenharia Química

17 / 11 / 21\nTHENDEFER, DOICI, GBR20191914\nlist Ψ:\n1) y''(t) - y(t) = 2t²\n y(0) = 1 , y'(0) = 0\n y(t) - yt) - 2t + t² = 0\n - APLICAR T1 NOS DOIS LADOS:\n L{y''(t)} - L{y(t)} = 2L{t²} = 2L{0} \n \n L{y''(t)} - a L{y(t)} - [y(0) - y(0)] = 0\n \n Y(s) = ... \n y(t) = 1 + E(t) - [y(t)]\n Y(s) = 5 + 5 + 2 - 2\ni + 1 + s\n y(t) = s⁴ + 2y - 2\n y(t) = \\frac{5 + 2 + 1}{5}x - (y^3)\n L^{-1} + (...) + 1 }(t) \n y(t) = ̶{y(t)} + d + v \n g? 15 / 11 / 21\n L^{-1} {s^5 - s^2 - 2y^2 + 1}\n s = 1\n L^{-1}(5 + b - 3y)\n \n 5 + 25s^2 - 2y = ...\n s₁ = 1 \n s_2 = 2\n E(s) + 5 = 0\n p(π) = ?\n =>\n p/5 - 1s - 5y\n (1 - 2)-...\n y(t) = 2x + 1/s + 1\n + +\n = {1}{2}y^2 + \n + e^t+ 1e^{t}\n 2{s} 23 / 11 / 21\nb) y''(t) - 2y'(t) + y(t) = t\n y(0) = 1 y'(0) = 1\n y(t) - t\n - APLICAR T1 NOS DOIS LADOS:\n L{y''(t)} = 2L{y'(t)} + L{y(t)}\n L{y(0)} = 2L{y(0)} - y(0) \n L{y'(t)} = saL{y(t)} - y(0)\n\n L{y''(t)} - sa L{y(t)}\n L{y(t)} = L(S) \n - SUBSTITUINDO\n L^2{y(s)} - y(0) - 2L{y(0)} + L{y(t)} - \n - RESCREVENDO\n L{y(0)} + L{y(0)} = L{y(0)} + Y(s) - L{y(0)} = 1 + Y(s) - 1 = 0\n...\n - ISOLAR Y(s)\n Y(s) (s^2 - 2s + 1) = ω₀ + 1 - 2ω₀ + s²\n Y(s) = (s₀ + 1 + 1)/(s^2 - 2s + 1)\n Y(s) = ( ω₀ - 1 + 1 ) x \n s₀Y(s) = s₀(g₀) - 2ω₀ + 1\n L Y(t) = D^{-1} {Y(s)}\n y(t) = D^{-1} {(s + 1)(sx + 1)} 23 11 21\n\n↦ EXPANDIR FUNÇÕES EM FRAÇÕES PARCIAIS:\n\\[\\frac{z^3}{\\omega^2+1} = A + B + C + D\\\\ \n\\frac{\\omega^2(z^2 - 1)}{(\\omega^1 - 1)^2} + (\\omega^1}^{2})(\\omega - 1)\\ = (A + A_2)(B) + 1\\\\ \np(\\infty) = 0,\ p(\\infty) = 1\\\\ B = 1\\ D = 1\\\\ -2A - 2B - C = -1 + C + 1\\\\ A = 2\\\\ \\ -4 + C = -1 \\\\ -C = -1 + 2\\ G=1.\\\\ Y_{d_{2} +\\leavevmode \\,} = Y(z),\\\\\n\\frac{1}{(\\omega-1)^2} \\\\ = 2.\\frac{z^{2}+1}{(\\omega^3 - 1)}\\\\\n(\\frac{1 + z^{1}}{(\\omega-1)^2})\\\\\n+ 1 + C)\\\\ y(t) = 2\\ + t - e^t + e^t 23 11 21\n\nC)\\ if\\ y^{(t)-2y^{(t)) + 5y(t) - 8e^{-t}\\ = y(0) = 2\\ y^{(0)} = 12\\\\ \n\\psi(t) - 2y(1) + 5y(t) + 0e^{t} = 0\\\\\nL^{3}y(t) - 2L^{3}{y(2)} + 5(L) + 0L^{(t)}=0 \\\\\nЕсли\\ L = y(0)= Y(s)\\\\\\\n\\ L\\ \\to Y(8)+ y(0) - 2L(0)\\ Y (0)+ 5y(t)\\ + 8\\\\\n\\ is end Y(5)\\\\\nY(0) = 2 + 10 - 4 - 8\\\\\nY(s)=\\frac{2^{3} + 10^{2}}{s^3 - \\cdots + 5\\\\\n - L^{-1}(\\frac{Y(0)}{(2^{3} + 10z)) = \\frac{z^{2} + 10^{2}}{(s^1 - 2^2-3^2 + 5}} 23 11 21\n\n↦ EXPANDIR.\n\\[\\frac{z^2}{s^2+10s} = K + \\frac{A_1}{s} + \\frac{A_2}{s^2+5}\\\\ \n\\frac{z^2+5}{s^2} = \\frac{1}{s(\\frac{\\omega_1}{s^2}+\\frac{\\omega+1}{s^2}+5)}\\\\ \nK = \\frac{a_2z^2+10z}{s^2+5}\\\\\n\\text{(s_0 + \\omega_0^2)(s_0 + \\omega_0+5)}\\\\ \nK = - 1\ \\\\ \n(\\omega_0^2 - 2\\omega_0 + 5)(\\omega_0) = -1\\\\ \n\\ -\\omega_2^2 + \\omega_1 = 5 + (\\omega + 1)(A + B) = -\\omega_2^2 + 5 + A\\omega_0 + A_2\\\\ \n\\frac{\\omega^2}{(\\omega+1)(\\omega_0^3 - 5\\omega^2 + A) + B - 5}\\\\\nA-1=2,\ A=3,\\ -1 + 3\\omega + 5\\\\\nA + B = 10 \\\\ B = 5\\\\\n-1 + 3(\\omega - 1) + B = 1,\\\\\n\\frac{z}{0^3+1} + 3.\\frac{z}{(\\omega - 1)^4}\\\\\nY(y)(t) = e^{-2t} + 3 e^{t} \\cos(zt) + 4 e^{t} \\sin(zt) 23 / 11 / 21\ndiy(t)−2y''(t)+y(t)=cos(3t) \ny(0)=0 \ny'(0)=1\nL[y''(t)]−2L[y'(t)]+L[y(t)]=L[cos(3t)]\n\n∎ SUBSTITUTING\ns^2L[y(t)]−sy(0)−2sL[y'(t)−y(0)]+L[y(t)]=L[cos(3t)]\n\ns^2L[y(t)]=Y(s)\n\n∎ PRESERVING\ns^2L[y(t)]−2sL[Y(s)]+Y(s)=\n Y(s) = Y(s)\ns^2-2s+1=0\n\n∎ ISOLATE, Y(s)\nY(0)=2+8\nd^{\omega^2}+g\n\nY(s)=L^{-1}{s^2+9+\frac{(s_0-1)^2}{(s_0+\omega^2)(s_0-1)^2}}=\ \nL^{-1}\{\frac{9}{s_0^2+9}(s_0-1)(s_0+g)\} \n\n∎ EXPAND\nY(s)=A_0R + B + C + D \n (s_0^2+9)(s_0-1)(s_0-1)^2\n\n 23 / 11 / 21\ns^2+g+9+(s_0-1)^2A_y(s) + B + (s_0^2+9)(s_0-1)e^{-(s_0-1)}C \n+\frac{(s_0^2+9}{(s_0-1)^2}D =\n(1+10) \n\n1 = 10 >\nd=10\n\nD = 11\n7.50 \n10\n-9 + 99 / 10\n\ns^2 + s \n+\ \frac{(s_0^2+9}{(s_0-1)^2}\n=1+\{(s_0^2+9) \\n+\hspace{2} 10G\}\n+g\}\n\ng=0 \n=> -g+99=10\n\nA + C=0\n-2A+B+C+11=1\n\n10\nC=-2\nA-2B+9C=-1\nB-9 + 99 = 9\n10 \n\n(y)\nY(s)= - \left( \frac{2}{25} s^2(t) - \frac{g}{50} + 2 +11 \right)\n+\frac{2}{(s+g)}*\frac{s(2)}{25}\n(2.5)(5)(10(t-1)^2)\n\n