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