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