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Question Number 103014 by bobhans last updated on 12/Jul/20

y′′ −y = (e^x /(e^x + e^(−x) )) .

$${y}''\:−{y}\:=\:\frac{{e}^{{x}} }{{e}^{{x}} \:+\:{e}^{−{x}} }\:. \\ $$

Answered by bobhans last updated on 12/Jul/20

the characteristic equation of the homogenous equation is γ^2 −1=0 ; γ=±1 y_h = C_1 e^x + C_2 e^(−x) say y_1 (x)=e^x →y_1 ′ = e^x y_2 (x)= e^(−x) →y_2 ′ = −e^(−x) Wronskian W(y_1 ,y_2 ) = determinant (((y_1 y_2 )),((y_1 ′ y_2 ′)))= determinant (((e^x e^(−x) )),((e^x −e^(−x) )))=−2≠0 I_1 = ∫ ((e^x ((e^x /(e^x +e^(−x) ))))/(W(y_1 ,y_2 ))) dx = −(1/2)∫ (e^(2x) /(e^x +e^(−x) )) dx = −(1/2)e^x +(1/2)tan^(−1) (e^x ) I_2 = ∫ ((e^(−x) ((e^x /(e^x +e^(−x) ))))/(−2)) dx = −(1/2)∫ (dx/(e^x +e^(−x) )) = −(1/2)tan^(−1) (e^x ) y_p = −e^x I_2 + e^(−x) I_1 = (1/2)e^x tan^(−1) (e^x )− (1/2)+(1/2)e^(−x) tan^(−1) (e^x ) complete solution y = C_1 e^x +C_2 e^(−x) +(1/2)(e^x + e^(−x) )tan^(−1) (e^x )−(1/2)

$${the}\:{characteristic}\:{equation}\:{of}\:{the} \\ $$$${homogenous}\:{equation}\:{is}\:\gamma^{\mathrm{2}} −\mathrm{1}=\mathrm{0}\:;\:\gamma=\pm\mathrm{1} \\ $$$${y}_{{h}} \:=\:{C}_{\mathrm{1}} {e}^{{x}} \:+\:{C}_{\mathrm{2}} {e}^{−{x}} \: \\ $$$${say}\:{y}_{\mathrm{1}} \left({x}\right)={e}^{{x}} \:\rightarrow{y}_{\mathrm{1}} '\:=\:{e}^{{x}} \\ $$$${y}_{\mathrm{2}} \left({x}\right)=\:{e}^{−{x}} \:\rightarrow{y}_{\mathrm{2}} '\:=\:−{e}^{−{x}} \\ $$$$\mathcal{W}{ronskian}\:\mathcal{W}\left({y}_{\mathrm{1}} ,{y}_{\mathrm{2}} \right)\:=\:\begin{vmatrix}{{y}_{\mathrm{1}} \:\:\:{y}_{\mathrm{2}} }\\{{y}_{\mathrm{1}} '\:\:{y}_{\mathrm{2}} '}\end{vmatrix}= \\ $$$$\begin{vmatrix}{{e}^{{x}} \:\:\:\:\:{e}^{−{x}} }\\{{e}^{{x}} \:\:−{e}^{−{x}} }\end{vmatrix}=−\mathrm{2}\neq\mathrm{0} \\ $$$${I}_{\mathrm{1}} =\:\int\:\frac{{e}^{{x}} \left(\frac{{e}^{{x}} }{{e}^{{x}} +{e}^{−{x}} }\right)}{\mathcal{W}\left({y}_{\mathrm{1}} ,{y}_{\mathrm{2}} \right)}\:{dx}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{{e}^{\mathrm{2}{x}} }{{e}^{{x}} +{e}^{−{x}} }\:{dx}\: \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{2}}{e}^{{x}} \:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left({e}^{{x}} \right) \\ $$$${I}_{\mathrm{2}} =\:\int\:\frac{{e}^{−{x}} \left(\frac{{e}^{{x}} }{{e}^{{x}} +{e}^{−{x}} }\right)}{−\mathrm{2}}\:{dx}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{{dx}}{{e}^{{x}} +{e}^{−{x}} } \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left({e}^{{x}} \right)\: \\ $$$${y}_{{p}} \:=\:−{e}^{{x}} {I}_{\mathrm{2}} +\:{e}^{−{x}} {I}_{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}{e}^{{x}} \:\mathrm{tan}^{−\mathrm{1}} \left({e}^{{x}} \right)− \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{x}} \:\mathrm{tan}^{−\mathrm{1}} \left({e}^{{x}} \right)\: \\ $$$${complete}\:{solution}\: \\ $$$${y}\:=\:{C}_{\mathrm{1}} {e}^{{x}} +{C}_{\mathrm{2}} {e}^{−{x}} +\frac{\mathrm{1}}{\mathrm{2}}\left({e}^{{x}} \:+\:{e}^{−{x}} \right)\mathrm{tan}^{−\mathrm{1}} \left({e}^{{x}} \right)−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by bramlex last updated on 12/Jul/20

coll joss

$$\mathrm{coll}\:\mathrm{joss}\: \\ $$

Answered by mathmax by abdo last updated on 12/Jul/20

y^(′′) −y =(e^x /(e^x +e^(−x) )) (he)→y^(′′) −y =0 →r^2 −1=0→r =+^− 1 ⇒y_h =a e^x +be^(−x) =au_1 +bu_2 W(u_1 ,u_2 ) = determinant (((e^x e^(−x) )),((e^x −e^(−x) ))) =−2 ≠0 W_1 = determinant (((o e^(−x) )),(((e^x /(e^x +e^(−x) )) −e^(−x) )))=−(1/(e^x +e^(−x) )) W_2 = determinant (((e^x 0)),((e^x (e^x /(e^x +e^(−x) ))))) =(e^(2x) /(e^x +e^(−x) )) v_1 =∫ (W_1 /W)dx =∫ (dx/(2(e^x +e^(−x) ))) =_(e^x =t) (1/2) ∫ (dt/(t(t +t^(−1) ))) =(1/2) ∫ (dt/(t^2 +1)) =(1/2)arctan(t) =(1/2)arctan(e^x ) v_2 =∫ (W_2 /W)dx =−∫ (e^(2x) /(2( e^x +e^(−x) )))dx =_(e^x =t) −(1/2) ∫ (t^2 /(t(t+t^(−1) ))) dt =−(1/2) ∫ (t/(t+t^(−1) ))dt =−(1/2)∫ (t^2 /(t^2 +1))dx =−(1/2)∫ ((t^2 +1−1)/(t^2 +1))dt =−(1/2)t +(1/2)arctan(t) =−(e^x /2) +(1/2) arctan(e^x ) ⇒ y_p =u_1 v_1 +u_2 v_2 =e^x ((1/2) arctan(e^x ))+e^(−x) (−(e^x /2) +(1/2) arctan(e^x )) =(e^x /2) arctan(e^x )−(1/2) +(e^(−x) /2)arctan(e^x ) =ch(x)arctan(e^x )−(1/2) the general solution is y =y_h +y_p =ae^x +be^(−x) +ch(x)arctan(e^x )−(1/2)

$$\mathrm{y}^{''} −\mathrm{y}\:=\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{e}^{\mathrm{x}} \:+\mathrm{e}^{−\mathrm{x}} } \\ $$$$\left(\mathrm{he}\right)\rightarrow\mathrm{y}^{''} −\mathrm{y}\:=\mathrm{0}\:\rightarrow\mathrm{r}^{\mathrm{2}} −\mathrm{1}=\mathrm{0}\rightarrow\mathrm{r}\:=\overset{−} {+}\mathrm{1}\:\Rightarrow\mathrm{y}_{\mathrm{h}} =\mathrm{a}\:\mathrm{e}^{\mathrm{x}} \:+\mathrm{be}^{−\mathrm{x}} \:=\mathrm{au}_{\mathrm{1}} \:+\mathrm{bu}_{\mathrm{2}} \\ $$$$\mathrm{W}\left(\mathrm{u}_{\mathrm{1}} ,\mathrm{u}_{\mathrm{2}} \right)\:=\begin{vmatrix}{\mathrm{e}^{\mathrm{x}} \:\:\:\:\:\:\:\:\:\:\mathrm{e}^{−\mathrm{x}} }\\{\mathrm{e}^{\mathrm{x}} \:\:\:\:\:\:\:\:\:−\mathrm{e}^{−\mathrm{x}} }\end{vmatrix}\:=−\mathrm{2}\:\neq\mathrm{0} \\ $$$$\mathrm{W}_{\mathrm{1}} =\begin{vmatrix}{\mathrm{o}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{e}^{−\mathrm{x}} }\\{\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{e}^{\mathrm{x}} \:+\mathrm{e}^{−\mathrm{x}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{e}^{−\mathrm{x}} }\end{vmatrix}=−\frac{\mathrm{1}}{\mathrm{e}^{\mathrm{x}} \:+\mathrm{e}^{−\mathrm{x}} } \\ $$$$\mathrm{W}_{\mathrm{2}} =\begin{vmatrix}{\mathrm{e}^{\mathrm{x}} \:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{e}^{\mathrm{x}} \:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{e}^{\mathrm{x}} \:+\mathrm{e}^{−\mathrm{x}} }}\end{vmatrix}\:=\frac{\mathrm{e}^{\mathrm{2x}} }{\mathrm{e}^{\mathrm{x}} \:+\mathrm{e}^{−\mathrm{x}} } \\ $$$$\mathrm{v}_{\mathrm{1}} =\int\:\frac{\mathrm{W}_{\mathrm{1}} }{\mathrm{W}}\mathrm{dx}\:=\int\:\:\frac{\mathrm{dx}}{\mathrm{2}\left(\mathrm{e}^{\mathrm{x}} \:+\mathrm{e}^{−\mathrm{x}} \right)}\:\:=_{\mathrm{e}^{\mathrm{x}} =\mathrm{t}} \:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\:\:\frac{\mathrm{dt}}{\mathrm{t}\left(\mathrm{t}\:+\mathrm{t}^{−\mathrm{1}} \right)}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\left(\mathrm{t}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\left(\mathrm{e}^{\mathrm{x}} \right) \\ $$$$\mathrm{v}_{\mathrm{2}} =\int\:\frac{\mathrm{W}_{\mathrm{2}} }{\mathrm{W}}\mathrm{dx}\:\:=−\int\:\:\frac{\mathrm{e}^{\mathrm{2x}} }{\mathrm{2}\left(\:\mathrm{e}^{\mathrm{x}} \:+\mathrm{e}^{−\mathrm{x}} \right)}\mathrm{dx}\:=_{\mathrm{e}^{\mathrm{x}} =\mathrm{t}} \:\:−\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{t}\left(\mathrm{t}+\mathrm{t}^{−\mathrm{1}} \right)}\:\mathrm{dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\frac{\mathrm{t}}{\mathrm{t}+\mathrm{t}^{−\mathrm{1}} }\mathrm{dt}\:=−\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx}\:=−\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{t}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{t}\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\left(\mathrm{t}\right)\:=−\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{arctan}\left(\mathrm{e}^{\mathrm{x}} \right)\:\Rightarrow \\ $$$$\mathrm{y}_{\mathrm{p}} =\mathrm{u}_{\mathrm{1}} \mathrm{v}_{\mathrm{1}} \:+\mathrm{u}_{\mathrm{2}} \mathrm{v}_{\mathrm{2}} =\mathrm{e}^{\mathrm{x}} \left(\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{arctan}\left(\mathrm{e}^{\mathrm{x}} \right)\right)+\mathrm{e}^{−\mathrm{x}} \left(−\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{arctan}\left(\mathrm{e}^{\mathrm{x}} \right)\right) \\ $$$$=\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{2}}\:\mathrm{arctan}\left(\mathrm{e}^{\mathrm{x}} \right)−\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{e}^{−\mathrm{x}} }{\mathrm{2}}\mathrm{arctan}\left(\mathrm{e}^{\mathrm{x}} \right) \\ $$$$=\mathrm{ch}\left(\mathrm{x}\right)\mathrm{arctan}\left(\mathrm{e}^{\mathrm{x}} \right)−\frac{\mathrm{1}}{\mathrm{2}}\:\:\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\:\:\mathrm{is} \\ $$$$\mathrm{y}\:=\mathrm{y}_{\mathrm{h}} \:+\mathrm{y}_{\mathrm{p}} =\mathrm{ae}^{\mathrm{x}} \:+\mathrm{be}^{−\mathrm{x}} \:+\mathrm{ch}\left(\mathrm{x}\right)\mathrm{arctan}\left(\mathrm{e}^{\mathrm{x}} \right)−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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