Question Number 103009 by bobhans last updated on 12/Jul/20

(D^2 −4D+4)y = x^3 e^(2x)

Answered by bramlex last updated on 12/Jul/20

since y = e^(2x)  is a part of the  homogenous solution (from  solving the characteristic  equation χ^2 −4χ+4 = 0 =  (χ−2)^2 =0. we can use  reduction of order . set y = z.e^(2x)   differentiating yields  Dy=(z′+2z)e^(2x)  and D^2 y =  (z′′+4z′+4z)e^(2x) .then  substituting into the original  differential equation yields   e^(2x) (z′′+4z′+4z)−4e^(2x) (z′+2z)+4z.e^(2x) =x^3   e^(2x)  which simplifies yields z′′=x^3   intgrating twice in succession  yields y = (C_1 x+C_2 )e^(2x)  + (1/(20))  x^5 e^(2x)  . hence a particular  solution is y = (1/(20))x^5 e^(2x)  . ⊕

Commented bybobhans last updated on 12/Jul/20

yes...

Answered by mathmax by abdo last updated on 12/Jul/20

y^(′′) −4y^′  +4 =x^3  e^(2x)   (he)→r^2 −4r +4 =0 ⇒(r−2)^2  =0⇒r=2 ⇒y_h =(ax+b)e^(2x)   =axe^(2x)  +be^(2x)  =au_1  +bu_2   W(u_1  ,u_2 ) = determinant (((xe^(2x  )             e^(2x) )),(((2x+1)e^(2x)   2e^(2x) )))=2xe^(4x) −(2x+1)e^(4x)  =−e^(4x)   ≠0  w_1 = determinant (((o          e^(2x) )),((x^3  e^(2x)     2e^(2x) )))=−x^3  e^(4x)   W_2 = determinant (((xe^(2x)                0)),(((2x+1)e^(2x)    x^3  e^(2x) )))=x^4  e^(4x)   v_1 =∫  ((W1)/W)dx =∫ ((−x^3  e^(4x) )/(−e^(4x) )) =∫ x^3  dx =(x^4 /4)  v_2 =∫ (W_2 /W)dx =∫  ((x^4  e^(4x) )/(−e^(4x) )) =−∫  x^4  dx =−(x^5 /5) ⇒  y_p =u_1 v_1  +u_2 v_2 =xe^(2x) ×(x^4 /4) +e^(2x) (−(x^5 /5)) =(x^5 /4) e^(2x) −(x^5 /5) e^(2x)   =(x^5 /(20)) e^(2x)   ⇒the general solution is  y =y_h  +y_p =(ax+b)e^(2x)  +(x^5 /(20)) e^(2x)

Commented bybemath last updated on 12/Jul/20

sir Abdo & sir Bramlex coll

Commented bymathmax by abdo last updated on 13/Jul/20

thanks sir