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【General guidance】The answer provided below has been developed in a clear step by step manner.Step1/3(a) In the given differential equation \( \mathrm{{p}{\left({t}\right)}=\frac{{1}}{{{2}{t}}},\quad{q}{\left({t}\right)}=\frac{{1}}{{{4}{t}}}} \)both these functions are not analytic at \( \mathrm{{t}={0}} \)But the functions \( \mathrm{{t}{p}{\left({t}\right)}{\quad\text{and}\quad}{t}^{{2}}{q}{\left({t}\right)}} \) are analytic at \( \mathrm{{t}={0}} \).Hence \( \mathrm{{t}={0}} \) is a regular singular point of the ODE.Explanation:Please refer to solution in this step.Step2/3(b) The Frobenius method:\( \mathrm{{y}={\sum_{{{n}={0}}}^{\infty}}{a}_{{n}}{t}^{{{n}+{r}}}} \)\( \mathrm{{y}'={\sum_{{{n}={0}}}^{\infty}}{a}_{{n}}{\left({n}+{r}\right)}{t}^{{{n}+{r}-{1}}}} \)\( \mathrm{{y}{''}={\sum_{{{n}={0}}}^{\infty}}{a}_{{n}}{\left({n}+{r}\right)}{\left({n}+{r}-{1}\right)}{t}^{{{n}+{r}-{2}}}} \)Substituting these in the differential equation:\( \mathrm{{4}{\sum_{{{n}={0}}}^{\infty}}{a}_{{n}}{\left({n}+{r}\right)}{\left({n}+{r}-{1}\right)}{t}^{{{n}+{r}-{1}}}+{2}{\sum_{{{n}={0}}}^{\infty}}{a}_{{n}}{\left({n}+{r}\right)}{t}^{{{n}+{r}-{1}}}+{\sum_{{{n}={0}}}^{\infty}}{a}_{{n}}{t}^{{{n}+{r}}}={0}} \)\( \mathrm{{\left[{4}{a}_{{0}}{r}{\left({r}-{1}\right)}+{2}{a}_{{0}}{r}\right]}{t}^{{{r}-{1}}}+{\sum_{{{n}={0}}}^{\infty}}{t}^{{{n}+{r}}}{\left[{a}_{{n}}+{4}{a}_{{{n}+{1}}}{\left({n}+{r}+{1}\right)}{\ ... See the full answer