Question Number 104595 by Ar Brandon last updated on 22/Jul/20

Solve the DE  x′′(t)+2x′(t)+x(t)=te^(−t)

Answered by mathmax by abdo last updated on 22/Jul/20

h)→r^2  +2r+1 =0 ⇒(r+1)^2  =0 ⇒r =−1 (double) ⇒x_h =(at+b)e^(−t)   =a t e^(−t)  +b e^(−t)  =au_1  +bu_2   W(u,,u_2 ) = determinant (((te^(−t)         e^(−t) )),(((1−t)e^(−t)   −e^(−t) )))=−te^(−2t) −(1−t)e^(−2t)  =(−t−1+t)e^(−2t)  =−e^(−2t) ≠0  W_1 = determinant (((o        e^(−t) )),((te^(−t)   −e^(−t) )))=−t e^(−2t)   W_2 = determinant (((te^(−t)       0)),(((1−t)e^(−t)        te^(−t) )))=t^2  e^(−2t)   v_1 =∫  (w_1 /W)dt =∫  ((−te^(−2t) )/(−e^(−2t) ))dt =∫t dt =(t^2 /2)  v_2 =∫ (w_2 /W)dt =∫  ((t^2  e^(−2t) )/(−e^(−2t) ))dt =−∫ t^2  dt =−(t^3 /3) ⇒  x_p =u_1 v_1  +u_2 v_2 =t e^(−t) ((t^2 /2))+e^(−t) (−(t^3 /3)) =((t^3 /2)−(t^3 /3))e^(−t)  =(t^3 /6)e^(−t)   the general solution is x =x_p  +x_h =(t^3 /6)e^(−t)  +(at+b)e^(−t)