Question Number 103201 by bobhans last updated on 13/Jul/20

(D^2 −2D+1)y = x ln(x)

Commented bybobhans last updated on 14/Jul/20

thank you both

Answered by bramlex last updated on 14/Jul/20

homogenous solution   η^2 −2η+1 = 0 ; (η−1)^2 = 0  y_h = C_1 e^x +C_2 xe^x    now we use variation of  parameters y=u.e^x  +v.xe^x   differentiating yields  y′=u′e^x +ue^x +v′xe^x +v(x+1)e^x   setting u′e^x +v′xe^x  = 0  y′′  =u′e^x +ue^x +v′(x+1)e^x +v(x+2)e^x   substituting y,y′,y′′ into original  differential equation gives us  (u′e^x +ue^x +v′(x+1)e^x +v(x+2)e^x )  −2(ue^x +v(x+1)e^x )+(ue^x +vxe^x )= xln(x)  which simplifies to   u′+(x+1)v′=xe^(−x) ln(x)  now we have two equations for  u′ and v′ ⇒u′e^x +v′xe^x =0  and u′+(x+1)v′ = xe^(−x) ln(x)  solving this system of equation  u′=−x^2 e^(−x) ln(x) and  v′=xe^(−x) ln(x) .integrating gives  us u= −∫_1 ^x t^2 e^(−t) ln(t)dt and   v = ∫_1 ^x  te^(−t)  ln(t)dt   hence we conclude that a  general solution is given by   y= C_1 e^x +C_2 xe^x −e^x ∫_1 ^x t^2 e^(−t) ln(t)dt+xe^x ∫_1 ^x te^(−t) ln(t)dt

Answered by abdomsup last updated on 13/Jul/20

(he)→y^(′′) −2y^′  +y =0 ⇒  r^2 −2r+1=0 ⇒(r−1)^2 =0 ⇒r=1  ⇒y_h =(ax+b)e^(x )  =axe^(x )  +be^x   =au_1 +bu_2   W(u_1  ,u_2 )= determinant (((xe^x          e^x )),(((x+1)e^x   e^x )))  =xe^(2x) −(x+1)e^(2x)   =−e^(2x)  ≠0  w_1 = determinant (((0        e^x )),((xlnx  e^x )))=−xe^x lnx  w_2 = determinant (((xe^x               o)),(((x+1)e^x      xlnx)))=  =x^2 e^x lnx  v_1 =∫(w_1 /w)dx =∫  ((xe^x lnx)/e^(2x) ) dx  =∫ x e^(−x) lnx dx  v_2 =∫ (w_2 /w)dx =−∫ ((x^2  e^x lnx)/e^(2x) )dx  =−∫ x^2  e^(−x) lnx dx  ⇒y_p =u_1 v_1  +u_2 v_2   =xe^x  ∫^x   t e^(−t) lnt dt−e^x  ∫^x  t^(2 ) e^(−t) lnt dt  y =y_h  +y_p