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【General guidance】The answer provided below has been developed in a clear step by step manner.Step1/2y''-2ty'+8y=0put y=\( \mathrm{\Sigma} \)an \( \mathrm{{t}^{{n}}} \)and y'=n\( \mathrm{\Sigma} \)an\( \mathrm{{t}^{{{n}-{1}}}} \)and y''=n(n-1)\( \mathrm{\Sigma{a}{n}{t}^{{{n}-{2}}}} \) put these values in IVPWe get,n(n-1)\( \mathrm{\Sigma{a}{n}{t}^{{{n}-{2}}}} \)-2t\( \mathrm{\Sigma{a}{n}{t}^{{{n}-{1}}}} \) +8\( \mathrm{\Sigma{a}{n}{t}^{{n}}} \)=0n(n-1)\( \mathrm{\Sigma{a}{n}{t}^{{{n}-{2}}}} \)-2\( \mathrm{\Sigma{a}{n}{t}^{{{n}}}} \) +8\( \mathrm{\Sigma{a}{n}{t}^{{n}}} \)=0n(n-1)\( \mathrm{\Sigma{a}{n}{t}^{{{n}-{2}}}=-{6}\Sigma{a}{n}{t}^{{n}}} \)n(n-1)\( \mathrm ... See the full answer