P2 - Cálculo 3 - 2024-1

1) f(t) = { e^3t , 0<=t<1 2 , 1<=t } L{f(t)} = ? L{f(t)} = ∫(0 to 1) f(t) e^-st dt + ∫(1 to ∞) 2 e^-st dt I II I ∫(0 to 1) e^(3-s)t dt = ∫(0 to 1) e^(t(3-s)) dt = [1/(3-s)] (e^t(3-s))|0 = 1/(3-s) (e^(3-s) - 1) = e^(3-s)/(3-s) - 1/(3-s) II II ∫(1 to ∞) 2 e^-st dt = -2 e^-st/s | 1 to ∞ = -2 (0 - e^-s)/s = 2 e^-s/s III L{f(t)} = e^(3-s)/(3-s) - 1/(3-s) + 2 e^-s/s I' = A * I 2) d/dt ( I ) = (0 V/L) ( I ) I : corrente ( V ) (-V/C -1/RC) ( V ) V : queda de voltagem a autovalores reais e iguais se L : 4R^2C A = (0 V/L) => (A - rI) ξ = 0 (-V/C -1/RC) det(A - rI) = 0 -r V/L r^2 + V/RC - (-V/LR^2C) = 0 -V/C -r - 1/RC ⟹ r + V/RC ± 4R^2C = 0 Δ = V^2/RC^2 -4.1. 4R^2C^2 = 0 => autovalores reais e iguais r1,2 = -1/RC = -1/2 2RC b R: 1ohm, C: 1farad e L: 4 Henry => sol. geral: ? r1,2 = -1/(2RC) = -1/2 (I^(1) / V(1)) = e^-rt ξ , solução Para r = -1/2 : ( 0 V/l ) ( ξ1 ) = ( 0 ) ⟹ ξ1 + ξ2 = 0 (-1 -1 ) ( ξ2 ) ( 0 ) 2 4 2ξ1 + ξ2 = 0 ∴ ( I^(1) ) = e^-t/2 ξ1 ( V^1 ) ξ1 = -2 ξ1 ⟹ ε̂(1) = ( 1 ) u(-2) ( -2 ) (I^(2) / V(2)) = c2[ (1 t.e^t/2) e^(V/4) (-2) => c0(5t e^rt, η e^rt) (A - rI) η = ξ (1/2 V/4) (η1) = (1 ) ⟹ η1 + η2 = 1 .1 -1/2) ( η2) = (-2 ) => 2η1 + η2 = 4 • η2 = 4 • η1 = 0 ⟹ η2 = 4 ⟹ η = (0) (4) (I/(V) ) = c0[ (1 t•e^1/2) e^v/4 + (0) (4) ∴ ( ( I) ) . c0[ (1 t.e^1/2) + c2 (1) e^(t/2) ( V ) (-2) (-2) t = 0: (1 ) = c0[(4) e^1/2) + c1 [ 0 ) e^0 + (0) e^e) (-2 (4)] 1 = c1 ⟹ c1 = 1 2 = -2c1 + 4c2 ⟹ 2 = 2 + 4c2 ⟹ c2 = 1 : ( I ) = [ (1) t•e^1/2 + (0) ( V ) ( (1) t•e^1/2 + c2(1) (-2) (-2) (4)] ] c I(t) e V(t): ? xe I(O) : 1A e V(O) = 2 vols ( (I) ) = c1[(1) e^(-t/2) + c2(1) (t)e^(t/2) + (0) (V) (-2) (-2) (4) t=0 : (1 ) = c1+(1)•e^(-0) + c2((0) e^(0) + (0) (-2) (-2)) (4) 1 = c1 + 0c2 => c1 = 1 2 = -2c1 + 4c2 => 2 = 2 + 4c2 => c2 = 1 ( ( I) ) .c1[ (1) t.e^(1/2) + (0) ( V ) (-2) (4)] bn = \frac{-2}{n\pi} \cos (\frac{n\pi}{2}) \frac{4}{n^2 \pi^2} \sen (\frac{n\pi}{2}) + \frac{4}{n\pi} [ \cos \frac{n\pi}{2} = (-1)^n ] \bn = \frac{2}{n\pi} \cos (\frac{n\pi}{2}) \frac{4}{n^2 \pi^2} \sen (\frac{n\pi}{2}) - \frac{4}{n\pi} (-1)^n \Para\ n\ impares:\ (n=2k+1): \cos (\frac{(2k+1)\pi}{2}) = 0, \ sen (\frac{(2k+1)\pi}{2}) = (.1)^k \therefore \, b_{2k+1} = \frac{4}{(2k+1)\pi^2} . (.1)^k = \frac{4}{(2k+1)\pi} . (-1)^k = \frac{4(.1)^k}{(2k+1)\pi} (\frac{1}{(2k+1)\pi} -1) \Para\ n\ pares:\ (n=2k) \cos \frac{2k\pi}{2} = (.1)^k , \ \sen \frac{2k\pi}{b} = 0 \therefore \, b_{2k} = \frac{1}{k\pi} . (.1)^k - \frac{2}{k\pi} (\frac{.1}{2k}) (.1)^k - \frac{2}{k\pi} (-.1)^k - \frac{2}{k\pi} = (.1)^k - \frac{2}{k\pi} \sp \spir\bullet (-1)^{2k} \equiv +1 \_____/_ I S T Q Q S S D 4) \ 3 ∂ 2 u = ∂ u I e ∂ 2 u = ∂ 2 u II ------------------- ------------------ ∂ x 2 ∂ x 2 ∂ y 2 ∂ y 2 I) 3 X'' - X Y'' => 3 X'' - X Y'' = 0 , A X'' + B X' Y' + C Y'' + D X Y + E X Y + F u = G Δ = B 2 - 4 A C , A = 3 , B = 0 e C = 0 Δ = -12 < 0 => Elíptica II) 3u. - 3u. = 0 , A:1 e C:-1 ------- --------- ∂x2 ∂y2 := Δ = 4 . 1 . (-1) = 4 > 0 => Hipérbola