TESTEBOLA 1653928393460 Cálculo III - 30-05 - Lista 02 pdf

1. (4+t^2) dy/dt - 2ty = 4t dy/dt - 2t/(4+t^2) y = 4t/(4+t^2) I(t) = e^(∫ -2t/(4+t^2) dt ) = e^(-ln|4+t^2|) = ln(1/(4+t^2)) I(t) = 1/(4+t^2) ∫ dy/4+t^2 - 2t/(4+t^2) dt = 4t/(4+t^2) (y/(4+t^2)) = 4t/(4+t^2)^2 ∫ 4t dt/(4+t^2) = -2/(4+t^2) + C y = -2 + C(4+t^2) 2. dy/dt + y/2 = e^(t/2) I(t)= exp/∫(1/2 dt) = exp(t/2) (e^(t/2) y)' = e^(5t/2) / 2 => (e^(1/2) y) = e^(5t/2) / 2 e^(t/2) y = ∫e^(5t/2) dt = e^(5t/2) * 6/5 + C y = 3e^(t/2) + Ce^(-t/2) 3. dy/dt = 4te^(t^2) - 3y y(0) = 2 y = 2y, p(t) = 4te^(t) y = -2y => 4te^(t^2) y = 4t e^(t^2) y = -2(e^(-2t)) + 2, y ) = , y ) 4. dy/dt = f(t) - y y(0) = , y = f(t) y'-1 => y = e^(t^2) - e^ 3. dy/dt = 4te^(t) - y y(0) = 1, y = 3e^(t) - y f(t), t^2 y(0) = 1=> y =, y = 3 , dy/dt = 3cos(2t) - y d(y(0) = y = 3), y 55 , f(t/y) y(0) = `=> y = e^(t^2) - f y(0) ==> y = 3 cos(2t)(et) y' -2 => 1. Solução crescem independentemente 2. Solução discutem independentemente 3. Solução analisam 4. Solução obtêm independentemente C. 1. y' + 3y = t + e^(-2t) I(t)= exp/∫ (3dt)= exp(3t) e^(3t) y = ∫e^(3t) * te^(2t) dt = y + e^3t + e^t y = 3 = e^(3y) 0 => y(∞) 2. y' - 2y = fe^(2t) I(t) = exp∫ (-2d•)= exp(-2t) e^( => ∫e^(2y)t = f•(t^ Y = -e^(2t)/(32) (8t + 4t^9) + Ce^3 3. y' + y = f•e^3 + I(t)= exp/(∫dt) e 4. y' + y = 3 cos(2(t)) = I(t/y), f'' = exp( => y Y = 3/4 ∫ - (9)(cos(2) y = 3/4, ∫τ(t) 14. y' + 2y = 2te^t Y(1) = 0 I(t) = e^2t => (e^2ty)'= 2te^3t => e^2ty = \int2te^3tdt = 2\intte^3tdt y = e^2t (4t - 1) + Ce^-2t y(3)=\frac{e^9.3 + Ce^2}{4} -\frac{36}{16} C = -\frac{24.3}{16} y(t) = e^2t (4t_t) - 3Ce^-2t 15. y' + \frac{y}{x} = \frac{1}{x} + 1 \nY(1) = \frac{1}{2} I(t) = x^2 \frac{2}{7} \int(x^2y) = \intx^2 * + x x^2y = \frac{x^4}{4} - \frac{x^3}{3} + \frac{x}{2} + C y = \frac{x^2}{4} + \frac{x}{2} + C y(1) = \frac{1}{4} = \frac{1_+}{4} + \frac{2}{3} + \frac{1}{2} => C = \frac{1}{4} - \frac{1_3}{4} = \frac{1}{12} y(x) = \frac{x^2}{4} - \frac{x}{4} + \frac{1}{2} + \frac{1}{8+12} 16. y' + \frac{x}{x^2} \cos(x) I(t) = x^2 = cos(x) y = \frac{\ln(x)}{x} + C n \ y(1) = 0 = 0 + \frac{C}{x^2} => C=0 x'(*t)= \frac{\ln(x)}{\frac{(x)}{11}} x = \frac{\ln(x)}{x} => \frac{x'}{x} = \intx^3d = \int\frac{(x'-x^3)}{x} => \int(4+x^3)^2 = \int(4x-x^3)^2dx => 4x+\frac{x^4}{1} = 2x^4 - 8x + 1 + C x^9+5x2 = 8x + \frac{16}{C'}C^4 = C+5 4\intx^xdx = 4 + x = \frac{1}{4} \intx^4dx x'(*t) = x^4 + C-\frac{C'}{x^3} => 4x + \frac{x^{2[t]}}{+t} + \frac{\intx^2(t)}{\intdt} x'(*5): y[$] + \frac{1}{L\frac{\frac{x*x^2*x}{(x*x)}}{3x-t}} +/- sqrt = cos(x) + c => y3 = cosinte cos(t) x = y(t) 9) 8. @ ds/dt = n s t = K => ds/dt - n s = k (e^(-nt)s) = K - ne^(nt) x t + C S(t) = Ce^nt - K/n S(0)=0 => C=K/n S(t) = Ke^nt/n - K/n Problem K = S/I n = 90% r = 40 K = 1500 => 3929,68 R$. e1 = 1 @ d ns = ke^nt + k nS = N s2 e^nt d s(t) n = 5000 500/(n+1500) = e^(40n) 25(40 n + 2/25) = e^(40n) 7(40 n + 8/25) e(-40n) = 2/25 - (40n + 2/25)e(-40n + 25) = - e^(-25) * (2/25) - (40n + 2/25) = W/(250 - w) => n = 1/40 n = 0.091154 = 9,11%. 9. S(t) = (e^nt + K/n) S(0) = 8000 n = 10% = 0,1 8000 = c + K => c = 8000 - 10K S(t) = (8000 - 10K) e^(0,1t) - 10K S(3) = 0 = (8000 - 10K) e^(0,3) - 10K 10K (8000 - 10K) e^(0,3) => 10K (1 + e^(0,3)) = 8000 e^(0,3) K = 800 e^(0,3) = 459,55 R$. t/e^(0,3t) 10. dy/dt = n(t) Y - k, Y(0) = y0 n(t) = (1/t + 1/n m(t))/5 @) dy/dt = nm(t) => Y(t) = -cos(t) + c Y0 = Y(0) = cos(0) + c => c = y0 + 1 Y(t) = y0 + 1 - cos(t) y0 = 1 => Y(t) = 3 - cos(t) 3/4 => Y(t) = 7/4 - cos(t) 1 => Y(t) = 2 - cos(t) d) X ≤ Y, x = y1 - cos(t) ≤ y0 + 2 y0 = 2 => y2 ≥ y2 e) dx/dt = (1/n nm(t) - K)/5 Y(t) = +/- sqrt(j - k)/5 - cos(t) + c Y(0)=y0 => C-J=y0 => C=y0+1 X(t) = t/5 - 5K - cos(t) + y0 + 1 -1 ≤ -cos(t) ≤ 1 Y0 ≤ Y + 1 - cos(t) ≤ y0 + 2 [+/- sqrt(1 - 5K)/5] 5 y0 + +/- sqrt(1 - 5K)/5 = y0 + 2 + 2 y0 + +/- sqrt(1 - 5K)/5 => y0 ≤ 2 +/- t/5 y0 ≤ 2 - t/5 -> 2 - 5K ∈ [-5K, +K/5] K=1/4 => y0 ≥ 2 + t/20 K=1/3 => y0 ≥ 2 + 2t/15 K=1/2 => y0 ≥ 2 + 3t/10 K=1/1 => y0 ≥ 2 + 4t/5 @) Graph K=3/4 K=1/2 y3 > 3/2 1/4 < 1/2 11. \frac{du}{dt} = -K(u-T(t)) T(t) = T_0 + T_1 \cos(\omega t) \indent \frac{du}{dt} + Ku = KT_0 + KT_1 \cos(\omega t) (e^{Kt}u)' = Ke^{Kt}[T_0 + T_1 \cos(\omega t)] e^{Kt}u = \int Ke^{Kt}[T_0 + T_1 \cos(\omega t)]dt + c = e^{Kt}\left[\frac{K}{K^2+\omega^2}\left(K\cos(\omega t)+\omega \sin(\omega t)\right) + \frac{T_0}{K}\right] + C u(t) = -\frac{K}{T_0}\left(K \cos(\omega t) + \omega \sin(\omega t)\right) + T_0 + Ae^{-Kt} S(t) = \frac{K}{T_0}\left(K \cos(\omega t) + \omega \sin(\omega t)\right) + T_0 + C e^{-Kt} (K+\omega^2) \omega = \frac{2\pi}{24}, T_0=60^\circ F, T_1=15^\circ F, k = \frac{\alpha}{c_x} S(t) = \frac{0.2}{1.5}0.98 \cos\left(\frac{2\pi}{24}\right) + \frac{0.2}{1.5} \sin\left(\frac{\pi}{24}\right) + 60 = \frac{0.4}{1.5 \sqrt{0.98}} T(t) = 60 + 15 \cos\left(\frac{2\pi t}{24}\right) \omega \approx 60 \theta Max(S) \approx 10^\circ F S \approx 50^\circ F \theta = 4 \to \text{Max(S)} \theta = 0 \to \text{Max(T)} \theta = 4 \text{ horas} www.codersil.com.br 12. \frac{dy}{dt} = \alpha y(1-y) \oint y=0 \to y=1 \oint \frac{dv}{dt} =\notin \underline{y=0} \to \text{initial} \underline{y=1} \to \text{statal} \frac{1}{y(1-y)} \cdot \frac{dy}{dt} = \alpha dt \int A \cdot \frac{1}{y} + B \cdot \frac{1}{1-y} = \frac{dy}{dt} \Rightarrow A(1-y) + By = 1 \frac{dy}{y} + \frac{dy}{1-y} = \alpha dt \ln|y| - \ln|1-y| = \alpha t + C \int \frac{dy}{y} = \alpha t + C \Rightarrow y = \frac{Ce^{\alpha t}}{1-y} \cdots y = \frac{Ce^{\alpha t}}{1+Ce^{\alpha t}} y = \frac{Ce^{\alpha t}}{1 + Ce^{\alpha t}}, t \rightarrow \infty \Rightarrow y \rightarrow 1 y = Ce^{\alpha t} \frac{1}{1+Ce^{\alpha t}} y(1+Ce^{\alpha t}) = Ce^{\alpha t}