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