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Question Number 97891 by pranesh last updated on 10/Jun/20

Commented by mr W last updated on 10/Jun/20

$$y=e^x-1$$

Commented by bemath last updated on 10/Jun/20

$${\lambda }^2-1=0$$$$\lambda =\pm 1$$$${\mathrm{y}}_{\mathrm{h}}={\mathrm{Ae}}^{\mathrm{x}}+{\mathrm{Be}}^{-\mathrm{x}}$$$$\mathrm{particular}{\mathrm{y}}_{\mathrm{p}}=ax+b$$$$y^{\prime }=a,{\mathrm{y}}^{″}=0$$$$0-(ax+b)=1\iff \{\begin{array}{l}a=0\\ b=-1\end{array}$$$$y_c=A{\mathrm{e}}^{\mathrm{x}}+{\mathrm{Be}}^{-\mathrm{x}}-1$$$$\mathrm{x}=0,\mathrm{y}=0$$$$\iff 0=\mathrm{A}+\mathrm{B}-1,\mathrm{A}+\mathrm{B}=1$$$${\mathrm{y}}_{\mathrm{c}}={\mathrm{Ae}}^{\mathrm{x}}+(1-\mathrm{A}){\mathrm{e}}^{-\mathrm{x}}-1$$

Commented by mr W last updated on 10/Jun/20

$$Amustbeequalto1,otherweise$$$$\underset{x\to -\mathrm{\infty }}{\lim }y=0+(1-A)\mathrm{\infty }-1=\mathrm{\infty }\ne a$$

Commented by bemath last updated on 10/Jun/20

$$\mathrm{how}\mathrm{to}\mathrm{get}\mathrm{A}=1?$$

Commented by mr W last updated on 10/Jun/20

$$suchthat\underset{x\to -\mathrm{\infty }}{\lim }y\nrightarrow \mathrm{\infty },theterm$$$$e^{-x}mustvanish,therefore1-A=0.$$

Commented by bemath last updated on 10/Jun/20

$$\mathrm{oo}\mathrm{yes}\mathrm{sir}.\mathrm{thank}\mathrm{you}$$

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