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

$$y^{″}-y=\frac{e^x}{e^x+e^{-x}}.$$

Answered by bobhans last updated on 12/Jul/20

$$thecharacteristicequationofthe$$$$homogenousequationis{\gamma }^2-1=0;\gamma =\pm 1$$$$y_h=C_1{e}^x+C_2{e}^{-x}$$$$say{y}_1\left(x\right)=e^x\to y_1^{\prime }=e^x$$$$y_2\left(x\right)=e^{-x}\to y_2^{\prime }=-e^{-x}$$$$\mathcal{W}ronskian\mathcal{W}(y_1,y_2)=\left|\begin{array}{c}{y}_1{y}_2\\ y_1^{\prime }{y}_2^{\prime }\end{array}\right|=$$$$\left|\begin{array}{c}{e}^x{e}^{-x}\\ e^x-e^{-x}\end{array}\right|=-2\ne 0$$$$I_1=\int \frac{e^x\left(\frac{e^x}{e^x+e^{-x}}\right)}{\mathcal{W}(y_1,y_2)}dx=-\frac{1}{2}\int \frac{e^{2x}}{e^x+e^{-x}}dx$$$$=-\frac{1}{2}{e}^x+\frac{1}{2}{\tan }^{-1}\left(e^x\right)$$$$I_2=\int \frac{e^{-x}\left(\frac{e^x}{e^x+e^{-x}}\right)}{-2}dx=-\frac{1}{2}\int \frac{dx}{e^x+e^{-x}}$$$$=-\frac{1}{2}{\tan }^{-1}\left(e^x\right)$$$$y_p=-e^x{I}_2+e^{-x}{I}_1=\frac{1}{2}{e}^x{\tan }^{-1}\left(e^x\right)-$$$$\frac{1}{2}+\frac{1}{2}{e}^{-x}{\tan }^{-1}\left(e^x\right)$$$$completesolution$$$$y=C_1{e}^x+C_2{e}^{-x}+\frac{1}{2}(e^x+e^{-x}){\tan }^{-1}\left(e^x\right)-\frac{1}{2}$$

Commented by bramlex last updated on 12/Jul/20

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

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

$${\mathrm{y}}^{{}^{″}}-\mathrm{y}=\frac{{\mathrm{e}}^{\mathrm{x}}}{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}$$$$\left(\mathrm{he}\right)\to {\mathrm{y}}^{{}^{″}}-\mathrm{y}=0\to {\mathrm{r}}^2-1=0\to \mathrm{r}=\stackrel{-}{+}1\Rightarrow {\mathrm{y}}_{\mathrm{h}}=\mathrm{a}{\mathrm{e}}^{\mathrm{x}}+{\mathrm{be}}^{-\mathrm{x}}={\mathrm{au}}_1+{\mathrm{bu}}_2$$$$\mathrm{W}({\mathrm{u}}_1,{\mathrm{u}}_2)=\left|\begin{array}{c}{\mathrm{e}}^{\mathrm{x}}{\mathrm{e}}^{-\mathrm{x}}\\ {\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}}\end{array}\right|=-2\ne 0$$$${\mathrm{W}}_1=\left|\begin{array}{c}\mathrm{o}{\mathrm{e}}^{-\mathrm{x}}\\ \frac{{\mathrm{e}}^{\mathrm{x}}}{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}-{\mathrm{e}}^{-\mathrm{x}}\end{array}\right|=-\frac{1}{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}$$$${\mathrm{W}}_2=\left|\begin{array}{c}{\mathrm{e}}^{\mathrm{x}}0\\ {\mathrm{e}}^{\mathrm{x}}\frac{{\mathrm{e}}^{\mathrm{x}}}{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}\end{array}\right|=\frac{{\mathrm{e}}^{2\mathrm{x}}}{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}$$$${\mathrm{v}}_1=\int \frac{{\mathrm{W}}_1}{\mathrm{W}}\mathrm{dx}=\int \frac{\mathrm{dx}}{2({\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}})}{=}_{{\mathrm{e}}^{\mathrm{x}}=\mathrm{t}}\frac{1}{2}\int \frac{\mathrm{dt}}{\mathrm{t}(\mathrm{t}+{\mathrm{t}}^{-1})}=\frac{1}{2}\int \frac{\mathrm{dt}}{{\mathrm{t}}^2+1}$$$$=\frac{1}{2}\arctan \left(\mathrm{t}\right)=\frac{1}{2}\arctan \left({\mathrm{e}}^{\mathrm{x}}\right)$$$${\mathrm{v}}_2=\int \frac{{\mathrm{W}}_2}{\mathrm{W}}\mathrm{dx}=-\int \frac{{\mathrm{e}}^{2\mathrm{x}}}{2({\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}})}\mathrm{dx}{=}_{{\mathrm{e}}^{\mathrm{x}}=\mathrm{t}}-\frac{1}{2}\int \frac{{\mathrm{t}}^2}{\mathrm{t}(\mathrm{t}+{\mathrm{t}}^{-1})}\mathrm{dt}$$$$=-\frac{1}{2}\int \frac{\mathrm{t}}{\mathrm{t}+{\mathrm{t}}^{-1}}\mathrm{dt}=-\frac{1}{2}\int \frac{{\mathrm{t}}^2}{{\mathrm{t}}^2+1}\mathrm{dx}=-\frac{1}{2}\int \frac{{\mathrm{t}}^2+1-1}{{\mathrm{t}}^2+1}\mathrm{dt}$$$$=-\frac{1}{2}\mathrm{t}+\frac{1}{2}\arctan \left(\mathrm{t}\right)=-\frac{{\mathrm{e}}^{\mathrm{x}}}{2}+\frac{1}{2}\arctan \left({\mathrm{e}}^{\mathrm{x}}\right)\Rightarrow$$$${\mathrm{y}}_{\mathrm{p}}={\mathrm{u}}_1{\mathrm{v}}_1+{\mathrm{u}}_2{\mathrm{v}}_2={\mathrm{e}}^{\mathrm{x}}\left(\frac{1}{2}\arctan \left({\mathrm{e}}^{\mathrm{x}}\right)\right)+{\mathrm{e}}^{-\mathrm{x}}(-\frac{{\mathrm{e}}^{\mathrm{x}}}{2}+\frac{1}{2}\arctan \left({\mathrm{e}}^{\mathrm{x}}\right))$$$$=\frac{{\mathrm{e}}^{\mathrm{x}}}{2}\arctan \left({\mathrm{e}}^{\mathrm{x}}\right)-\frac{1}{2}+\frac{{\mathrm{e}}^{-\mathrm{x}}}{2}\arctan \left({\mathrm{e}}^{\mathrm{x}}\right)$$$$=\mathrm{ch}\left(\mathrm{x}\right)\arctan \left({\mathrm{e}}^{\mathrm{x}}\right)-\frac{1}{2}\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{is}$$$$\mathrm{y}={\mathrm{y}}_{\mathrm{h}}+{\mathrm{y}}_{\mathrm{p}}={\mathrm{ae}}^{\mathrm{x}}+{\mathrm{be}}^{-\mathrm{x}}+\mathrm{ch}\left(\mathrm{x}\right)\arctan \left({\mathrm{e}}^{\mathrm{x}}\right)-\frac{1}{2}$$

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