Question Number 103203 by bobhans last updated on 13/Jul/20

y′′+4y′+4y = (e^(−2x) /x^2 )

Answered by bramlex last updated on 14/Jul/20

characteristic eq   χ^2 +4χ+4 = 0 ; (χ+2)^2 =0  y_h = C_1 e^(−2x) +C_2 xe^(−2x)   now we apply reduction of order  set y = e^(−2x) .z , substituting this  into the original differential  equation yields   e^(−2x) (z′′−4z′+4z)+e^(−2x) (z′−2z)  +4z e^(−2x)  = (e^(−2x) /x^2 ) . which  simplifies to z′′ = (1/x^2 )  intrgrating this twice succession  yields z = −ln∣x∣ +C_1 x+C_2   hence a general solution is given  by y = e^(−2x) {−ln∣x∣+C_2 x+C_1 }   y = C_1 e^(−2x) + C_2 x e^(−2x) −e^(−2x) ln∣x∣ □

Commented bybobhans last updated on 13/Jul/20

cooll....

Answered by mathmax by abdo last updated on 13/Jul/20

y^(′′)  +4y^′  +4y =(e^(−2x) /x^2 )  (he)→y^(′′)  +4y^′  +4y =0 →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) )),(((1−2x)e^(−2x)    −2e^(−2x) )))=−2x e^(−2x) +(2x−1)e^(−2x)  =−e^(−2x )  ≠0  W_1 = determinant (((o             e^(−2x) )),(((e^(−2x) /x^2 )         −2e^(−2x) )))=−(e^(−4x) /x^2 )  W_2 = determinant (((xe^(−2x)              0)),(((1−2x)e^(−2x)      (e^(−2x) /x^2 ))))=(e^(−4x) /x)  v_1 =∫ (w_1 /w)dx =∫ (e^(−4x) /(x^2  e^(−2x) )) dx = ∫  (e^(−2x) /x^2 ) dx  v_2 =∫ (w_2 /w) dx =−∫ (e^(−4x) /(x e^(−2x) ))dx =−∫ (e^(−2x) /x)dx ⇒  y_p =u_1 v_1  +u_2 v_2 =xe^(−2x)  ∫^x  (e^(−2t) /t)dt  −e^(−2x )  ∫^x   (e^(−2t) /t) dt  y =y_h  +y_p