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Question Number 2161 by Yozzis last updated on 05/Nov/15
$${Solve}\:{the}\:{d}.{e}\: \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }\sqrt{\frac{{dy}}{{dt}}}−\frac{{t}}{{y}}=\mathrm{0}. \\ $$
Answered by prakash jain last updated on 07/Nov/15
$$\mathrm{Solving}\:\mathrm{as}\:\mathrm{a}\:\mathrm{series}\:\mathrm{expansion} \\ $$$${y}=\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{i}} {t}^{{i}} \\ $$$$\frac{{dy}}{{dt}}=\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{ia}_{{i}} {t}^{{i}−\mathrm{1}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }=\underset{{i}=\mathrm{2}} {\overset{\infty} {\sum}}{i}\left({i}−\mathrm{1}\right){a}_{{i}} {t}^{{i}−\mathrm{2}} \\ $$$${y}^{\mathrm{2}} \left(\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }\right)^{\mathrm{2}} \left(\frac{{dy}}{{dt}}\right)={t}^{\mathrm{2}} \\ $$$$\left(\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{i}} {t}^{{i}−\mathrm{1}} \right)^{\mathrm{2}} \left(\underset{{i}=\mathrm{2}} {\overset{\infty} {\sum}}{i}\left({i}−\mathrm{1}\right){a}_{{i}} {t}^{{i}−\mathrm{2}} \right)^{\mathrm{2}} \left(\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{ia}_{{i}} {t}^{{i}−\mathrm{1}} \right)=\mathrm{1} \\ $$$${equating}\:{coefficients} \\ $$$$\left({t}^{\mathrm{0}} \right)\:\:\:\:{a}_{\mathrm{1}} \centerdot\mathrm{2}{a}_{\mathrm{2}} \centerdot{a}_{\mathrm{1}} =\mathrm{1}\:\: \\ $$$${more}\:{work}\:{to}\:{be}\:{done}. \\ $$