Question Number 105195 by mohammad17 last updated on 26/Jul/20

Answered by Aziztisffola last updated on 26/Jul/20

L(2y′′+3y′−2y)=L(te^(−2t) )  2s^2 Y(s)−2sy(0)−2y(0)+3sY(s)−3y(0)−2Y(s)=−(d/ds)(L(e^(−2t) ))  (2s^2 +3s−2)Y(s)+4=−(d/ds)((1/(s+2)))  (2s^2 +3s−2)Y(s)+4=(1/((s+2)^2 ))  Y(s)=((1/((s+2)^2 ))−4).(1/(2s^2 +3s−2))  =((1/((s+2)^2 ))−4).(1/((s+2)(s−(1/2))))  =(1/((s+2)^3 (s−(1/2)))) − (4/((s+2)(s−(1/2))))  then partial fractions and calulate the  inverse y(t)=L^(−1) {Y(s)}

Answered by mathmax by abdo last updated on 27/Jul/20

2y^(′′) +3y^′ −2y =te^(−2t)  ⇒2L(y^(′′) )+3L(y^′ )−2L(y) =L(te^(−2t) )  ⇒2{t^2 L(y)−ty(o)−y^′ (o))+3{tL(y)−y(o)}−2L(y)=L(te^(−2t) ) ⇒  (2t^2  +3t−2)L(y)+4 =L(te^(−2t) ) but  L(te^(−2t) ) =∫_0 ^∞  xe^(−2x)  e^(−tx) dx =∫_0 ^∞  x e^(−(t+2)x) dx  =[(x/(−(t+2)))e^(−(t+2)x) ]_(x=0) ^∞  −∫_0 ^∞ (1/(−(t+2)))e^(−(t+2)x) dx  =(1/((t+2)))∫_0 ^∞  e^(−(t+2)x) dx =−(1/((t+2)^2 ))[e^(−(t+2)x) ]_(x=0) ^∞  =(1/((t+2)^2 ))  e ⇒(2t^2 +3t−2)L(y) =−4 +(1/((t+2)^2 )) ⇒  L(y) =((−4)/(2t^2 +3t−2)) +(1/((t+2)^2 (2t^2 +3t−2))) ⇒  y(t) =−4L^(−1) ((1/(2t^2 +3t−2))) +L^(−1) ((1/((t+2)^2 (2t^2  +3t−2)))  let decompose F(t)=(1/(2t^2  +3t−2))  Δ =9−4(−4) =25 ⇒t_1 =((−3+5)/4) =(1/2) and t_2 =((−3−5)/4) =−2 ⇒  F(x) =(1/(2(t−t_1 )(t−t_2 ))) =(1/2)×(2/5)((1/(t−(1/2)))−(1/(t+2))) ⇒  L^(−1) (F(t)) =(1/5){ e^(x/2)  −e^(−2x) }  let decompose g(t)=(1/((t+2)^2 (2t^2  +3t−2)))  ⇒g(t) =(1/(2(t+2)^2 (t+2)(t−(1/2)))) =(a/(t+2)) +(b/((t+2)^2 )) +(c/((t+2)^3 )) +(d/(t−(1/2)))  ⇒L^(−1) (g) =a e^(−2t)  +bt e^(−2t)  +ct^2 e^(−2t)  +d e^(t/2)