Blog: Having trouble understanding how to get started on the attached problem in Asymptotics. Any help is greatly appreciated.

Sunday, December 30, 2012

Having trouble understanding how to get started on the attached problem in Asymptotics. Any help is greatly appreciated.

You need to write the characteristic equation, such that:

$\displaystyle{{r}}^{{2}}-{\left(\frac{{1}}{{2}}\right)}{r}+\frac{{1}}{{16}}+{x}-{{e}}^{{x}}={0}$

You need to evaluate $\displaystyle\delta$ , such that:

$\displaystyle\delta={{\left(-\frac{{1}}{{2}}\right)}}^{{2}}-{4}{\left(\frac{{1}}{{16}}+{x}-{{e}}^{{x}}\right)}=\frac{{1}}{{4}}-\frac{{1}}{{4}}-{4}{x}+{4}{{e}}^{{x}}$

$\displaystyle\delta={4}{{e}}^{{x}}-{4}{x}$

The characteristic equation has real solutions if $\displaystyle\delta>{0}$

$\displaystyle{4}{{e}}^{{x}}-{4}{x}>{0}\Rightarrow{{e}}^{{x}}-{x}>{0}\Rightarrow{{e}}^{{x}}>{x}$

Attaching the equation yields:

$\displaystyle{{e}}^{{x}}={x}\Rightarrow{{\ln{{e}}}}^{{x}}={\ln{{x}}}\Rightarrow{x}={\ln{{x}}}\Rightarrow\frac{{{\ln{{x}}}}}{{x}}={1}$

If $\displaystyle{x}\to\infty$ yields $\displaystyle\lim_{{{x}\to\infty}}\frac{{{\ln{{x}}}}}{{x}}={0}$

Hence, $\displaystyle{r}_{{{1},{2}}}=\frac{{\frac{{1}}{{2}}\pm\sqrt{{{4}{\left({{e}}^{{x}}-{x}\right)}}}}}{{2}}\Rightarrow{r}_{{{1},{2}}}=\frac{{\frac{{1}}{{2}}\pm{2}\sqrt{{{{e}}^{{x}}-{x}}}}}{{2}}\Rightarrow{r}_{{{1},{2}}}=\frac{{1}}{{4}}\pm\sqrt{{{{e}}^{{x}}-{x}}}$

If $\displaystyle{x}\to\infty$ yields $\displaystyle{r}_{{{1},{2}}}=\frac{{1}}{{4}}\pm\infty\Rightarrow{r}_{{{1},{2}}}=\pm\infty$

Hence, the general solution to the equation, for $\displaystyle{x}\to\infty$ , is:

$\displaystyle{y}{\left({t}\right)}={c}_{{1}}\cdot{{e}}^{{+\infty\cdot{t}}}+{c}_{{2}}\cdot{{e}}^{{-\infty\cdot{t}}}$

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Sunday, December 30, 2012

Having trouble understanding how to get started on the attached problem in Asymptotics. Any help is greatly appreciated.

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