Question Number 103478 by mohammad17 last updated on 15/Jul/20

Answered by bemath last updated on 15/Jul/20

HE ⇒λ^2 +3λ−2=0  (λ+(3/2))^2 −(9/4)−2=0  (λ+(3/2))^2 =((17)/4) ⇒λ=−(3/2)±((√(17))/2)  y_h =C_1 e^(((−3+(√(17)))x)/2) +C_2 e^(((−3−(√(17)))x)/2)   particular solution  y_p =−((11)/(101))sin 3t−(9/(101))cos 3t  general solution  y=C_1 e^(((−3+(√(17)))x)/2) +C_2 e^(((−3−(√(17)))x)/2) −  ((11)/(101))sin 3t−(9/(101))cos 3t  y(0) ⇒C_1 +C_2 =(9/(101))  y′(x)=(((−3+(√(17)))/2))C_1 e^(((−3+(√(17)))x)/2) −  C_2 (((3+(√(17)))/2))e^(((−3−(√(17)))x)/2)

Answered by Worm_Tail last updated on 15/Jul/20

(d^2 y/dt^2 )+3(dy/dt)−2y=2sin3t         L((d^2 y/dt^2 )+3(dy/dt)−2y)=L(2sin3t )     L((d^2 y/dt^2 ))+L(3(dy/dt))−L(2y)=(6/(s^2 +9))     s^2 F(s)−sy(0)−y′(0)+3(sF(s)−y(0))−2F(s)=(6/(s^2 +9))     s^2 F(s)−(−2)+3sF(s)−2F(s)=(6/(s^2 +9))     F(s)[s^2 +3s−2]=(6/(s^2 +9))−2     F(s)[s^2 +3s−2]=((−2s^2 −12)/(s^2 +9))     F(s)=((−2s^2 −12)/((s^2 +9)(s^2 +3s−2)))     F(s)=(9/(101))((s/((s−(3/2))^2 −((19)/4))))−((142)/(101))((1/((s−(3/2))^2 −((19)/4))))−(9/(101))((s/(s^2 +9)))−((33)/(101))((1/(s^2 +9)))  evaluating  L^(−1)  of both sides     f(t)=(e^((−t(3+(√(17))))/2) /(34(101)))(−311(√(17))e^(t(√(17)))   +153e^(t(√(17))) +153+311(√(17)))−(((9x+11)sin(3t))/(3(101)))

Answered by mathmax by abdo last updated on 15/Jul/20

y^(′′)  +3y^′ −2y =2sin(3t)  he→y^(′′) +3y^′ −2y =0→r^2 +3r−2=0 →Δ=9+8 =17 ⇒  r_1 =((−3+(√(17)))/2)  and r_2 =((−3−(√(17)))/2) ⇒y_h =ae^(r_1 t)  +b e^(r_2 t)  =au_1  +bu_2   W(u_1 ,u_2 ) = determinant (((e^(r_1 t)           e^(r_2 t) )),((r_1 e^(r_1 t)       r_2 e^(r_2 t) )))=(r_2 −r_1 )e^((r_1 +r_2 )t)  =−(√(17))e^(−3t)  ≠0  W_1 = determinant (((o          e^(r_2 t) )),((2sin(3t)    r_2 e^(r_2 t) )))=−2e^(r_2 t)  sin(3t)  W_2 = determinant (((e^(r_1 t)            0)),((r_1 e^(r_1 t)      2sin(3t))))=2e^(r1t)  sin(3t)  v_1 =∫ (w_1 /w)dt =−∫  ((2e^(r_2 t) sin(3t))/(−(√(17))e^(−3t) )) =(2/(√(17))) ∫ e^((3+r_2 )t)  sin(3t)dt  =(2/(√(17))) ∫ e^((((3−(√(17)))/2))t)  sin(3t)dt =(2/(√(17))) Im(∫ e^((((3−(√(17)))/2)+3i)t) dt)  and ∫  e^((...)t)  dt =(1/((((3−(√(17)))/2)+3i)))e^((((3−(√(17)))/2)+3i)t)   =(2/((3−(√(17))+6i)))e^((...))  =((2(3−(√(17))−6i))/((3−(√(17)))^2  +36)) e^(((3−(√(17)))/2)t) {cos(3t)+isin(3t)} rest  to extract Im(∫..)..  v_2 =∫ (w_2 /w)dt =∫ ((2e^(r_1 t)  sin(3t))/(−(√(17))e^(−3t) )) =−(2/(√(17))) ∫  e^((3+r_1 )t)  sin(3t)  =−(2/(√(17))) Im(∫  e^((3+r_1 +3i)t) dt) =....we follow the same way..  ⇒y_p =u_1 v_1  +u_2 v_2   and y =y_h  +y_p

Answered by mathmax by abdo last updated on 15/Jul/20

let use laplace transform  y^(′′)  +3y^′  −2y =2sin(3t) ⇒L(y^(′′) )+3L(y^′ )−2L(y) =2L(sin(3t)) ⇒  t^2 L(y)−ty(o)−y^′ (o)+3(t L(y)−y(o))−2L(y) =2L(sin(3t)) ⇒  (t^2 +3t−2)L(y) +2 =2 L(sin(3t)) and  L(sin(3t)) =∫_0 ^∞   sin(3x)e^(−tx)  dx  =Im(∫_0 ^∞  e^(−tx+3ix)  dx)   =Im(∫_0 ^∞  e^((−t+3i)x)  dx) and  ∫_0 ^∞  e^((−t+3i)x)  dx =[(1/(−t+3i)) e^((−t+3i)x) ]_0 ^∞   =−(1/(−t+3i)) =(1/(t−3i)) =((t+3i)/(t^2  +9)) ⇒L(sin(3t)) =(3/(t^2  +9))  e⇒ (t^2  +3t−2)L(y) =−2 +(6/(t^2  +9)) ⇒L(y) =((−2)/(t^2 +3t−2)) +(6/((t^2  +9)(t^2 +3t−2)))  ⇒y(t) =−2L^(−1) ((1/(t^2  +3t−2))) +6L^(−1) ((1/((t^2 +9)(t^2 +3t−2))))  let decompose f(t) =(1/(t^2 +3t−2))  Δ =9+8=17 ⇒t_1 =((−3+(√(17)))/2)  and t_2 =((−3−(√(17)))/2) ⇒  f(t)=(1/((t−t_1 )(t−t_2 ))) =(1/(√(17)))((1/(t−t_1 ))−(1/(t−t_2 )) ) ⇒  L^(−1) (f) =(1/(√(17)))e^(t_1 t) −(1/(√(17))) e^(t_2 t)   let decompose g(t) =(1/((t^2 +9)(t^2  +3t−2)))  ⇒g(t) =(1/((t−3i)(t+3i)(t−t_1 )(t−t_2 ))) =(a/(t−3i)) +(b/(t+3i)) +(c/(t−t_1 )) +(d/(t−t_2 ))  eazy to find this coefficients ⇒  L^(−1) (g) =a e^(3it)  +b e^(−3it)  +c e^(t_1 t)  +de^(t_2 t)   ⇒at form  αcos(3t) +βsin(3t) +ce^(t_1 t)  +d e^(t_2 t)  ⇒  y(t) =αcos(3t)+βsin(3t) +a e^((((−3+(√(17)))/2))t)  +b e^((((−3−(√(17)))/2))t)