Question Number 101419 by bemath last updated on 02/Jul/20

y′′−4y′+3y = (e^x /(1+e^x ))

Answered by john santu last updated on 02/Jul/20

AE: λ^2 −4λ+3 = 0 (λ−1)(λ−3)=0 y_h = C_1 e^x + C_2 e^(3x ) →y_1 = e^x ∧y_2 = e^(3x) W = determinant (((y_1 y_2 )),((y_1 ^′ y_2 ′)))= determinant (((e^x e^(3x) )),((e^x 3e^(3x) ))) W= 3e^(4x) −e^(4x) = 2e^(3x) particular integral =−(e^x /2)∫ (e^x /(e^x (1+e^x ))) dx + (e^(3x) /2)∫ (e^x /e^(4x) ) .(e^x /((1+e^x ))) dx =−(e^x /2)∫ (dx/(1+e^x )) + (e^(3x) /2)∫ (dx/(e^(2x) (1+e^x ))) I_1 = ∫ (dx/(1+e^x )) [ set e^x = z , dx = (dz/e^x )] I_1 = ∫ (dz/(z(1+z))) = ∫ (dz/z)−∫ (dz/(1+z)) I_1 = ln((z/(1+z))) = ln ((e^x /(1+e^x ))) I_2 = ∫(dx/(e^(2x) (1+e^x ))) [ set e^x = p ] I_2 = ∫ (dp/(p^3 (1+p))) = ∫ (dp/p)−∫ (dp/p^2 ) +∫ (dp/p^3 )−∫ (dp/((1+p))) I_2 = ln(p)+(1/p)−(1/(2p^2 ))−ln(1+p) I_2 = ln((e^x /(1+e^x ))) +(1/e^x ) −(1/(2e^(2x) )) particular solution y_p = −(e^x /2)ln((e^x /(1+e^x )))+(e^(3x) /2)((1/e^x )−(1/(2e^(2x) )) + ln((e^x /(1+e^x )))) Generall solution y_g = y_h + y_p

Answered by mathmax by abdo last updated on 02/Jul/20

(he)→y^(′′) −4y^′ +3y =0 →r^2 −4r +3=0 Δ^′ =4−3 =1 ⇒ r_1 =2+1 =3 and r_2 =2−1=1 ⇒y_h =ae^(3x) +be^x =au_1 +bu_2 W(u_1 ,u_2 ) = determinant (((e^(3x ) e^x )),((3e^(3x) e^x )))=e^(4x) −3e^(4x) =−2e^(4x) W_1 = determinant (((o e^x )),(((e^x /(1+e^x )) e^x )))=−(e^(2x) /(1+e^x )) W_2 = determinant (((e^(3x) 0)),((3e^(3x) (e^x /(1+e^x )))))=(e^(4x) /(1+e^x )) v_1 =∫ (w_1 /w)dx =∫ −(e^(2x) /(1+e^x ))×(1/(−2e^(4x) )) =∫ (e^(−2x) /(2(1+e^x ))) =_(e^x =t) (1/2)∫ (1/(t^2 (1+t)))×(dt/t) =(1/2)∫ (dt/(t^3 (1+t))) let decompose F(t) =(1/((t+1)t^3 )) F(t) =(a/(t+1)) +(b/t) +(c/t^2 ) +(d/t^3 ) a =−1 ,d =1 ⇒F(t) =−(1/(t+1)) +(b/t) +(c/t^2 ) +(1/t^3 ) lim_(t→+∞) tF(t) =0 =−1+b ⇒b=1 ⇒F(t) =−(1/(t+1)) +(1/t) +(c/t^2 ) +(1/t^3 ) F(1) =(1/2) =−(1/2) +1+c +1 =(3/2) +c ⇒c =(1/2)−(3/2) =−1 ⇒ F(t) =−(1/(t+1)) +(1/t)−(1/t^2 ) +(1/t^3 ) ⇒v_1 =(1/2)(−ln∣t+1∣+ln∣t∣+(1/t) −(1/(2t^2 ))) =(1/2)(ln((e^x /(1+e^x ))) +e^(−x) −(1/2)e^(−2x) ) v_2 =∫ (w_2 /w)dx =∫ (e^(4x) /((1+e^x )(−2e^(4x) )))dx =−(1/2)∫ (dx/(1+e^x )) =_(e^x =t) −(1/2)∫ (dt/(t(1+t))) =−(1/2)∫((1/t)−(1/(t+1)))dt =−(1/2)ln∣(t/(t+1))∣ =−(1/2)ln((e^x /(1+e^x ))) ⇒y_p =u_1 v_1 +u_2 v_2 =e^(3x) ×(1/2){ln((e^x /(1+e^x )))+e^(−x) −(1/2)e^(−2x) } +e^x ×(−(1/2)ln((e^x /(1+e^x )))) =(1/2) e^(3x) ln((e^x /(1+e^x )))+(1/2)e^(2x ) −(1/4) e^x −(1/2)e^x ln((e^x /(1+e^x ))) =(1/2)(e^(3x) −e^x )ln((e^x /(1+e^x )))+(e^(2x) /2)−(e^x /4) the general solution is y =y_h +y_p