Question Number 103995 by mathmax by abdo last updated on 18/Jul/20

solve y^(′′) +2y^′ −y =(e^(−x) /x)

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

h→r^2 +2r−1 =0 →Δ^′  =1+1=2 ⇒r_1 =−1+(√2) and r_2 =−1−(√2) ⇒  y_h =ae^((−1+(√2))x)  )+b e^((−1−(√2))x)   =au_1  +bu_2   W(u_1  ,u_2 ) = determinant (((e^((−1+(√2))x)                                      e^((−1−(√2))x) )),(((−1+(√2))e^((−1+(√2))x)    (−1−(√2)) e^((−1−(√2))x) )))  =(−1−(√2))e^(−2x) −(−1+(√2))e^(−2x)  =−2(√2)e^(−2x)   W_1 = determinant (((o           e^((−1−(√2))x) )),(((e^(−x) /x)        (−1−(√2))e^((−1−(√2))x) ))) =−(e^(−x) /x) e^((−1−(√2))x)  =−(1/x)e^((−2−(√2))x)   W_2 = determinant (((e^((−1+(√2))x)            0)),(((−1+(√2))e^x     (e^(−x) /x))))=(1/x) e^((−2+(√2))x)   v_1 =∫ (w_1 /w)dx =−∫  (e^((−2+(√2))x) /(−2(√2)x e^(−2x) ))dx =(1/(2(√2))) ∫  (e^((√2)x) /x)dx  v_2 =∫ (w_2 /w)dx =−(1/(2(√2)))∫  (e^((−2+(√2))x) /(x e^(−2x) ))dx =−(1/(2(√2)))∫  (e^((√2)x) /x)dx ⇒  y_p = u_1 v_1  +u_2 v_2 =(1/(2(√2)))e^((−1+(√2))x)    ∫ (e^(x(√2)) /x)dx −(1/(2(√2)))e^((−1−(√2))x) ∫ (e^(x(√2)) /x)dx  =(e^(−x) /(√2)){((e^(x(√2)) −e^(−x(√2)) )/2)}∫  (e^(x(√2)) /2)dx =(e^(−x) /(√2))sh(x(√2))∫  (e^(x(√2)) /x)dx  the general solution is y =y_h  +y_p   ⇒y =a e^((−1+(√2))x)  +b e^((−1−(√2))x)  + (e^(−x) /(√2))sh(x(√2))∫  (e^(x(√2)) /x)dx