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 $$\mathrm{Solving}\:\mathrm{as}\:\mathrm{a}\:\mathrm{series}\:\mathrm{expansion}…
Question Number 67697 by Rasheed.Sindhi last updated on 30/Aug/19 $$\Cup\mathrm{si}\Cap\mathrm{g}\:\mathrm{ChineseRemainderTheorm} \\ $$$$\partial\mathrm{etermine}\:\mathrm{polynomial}\:\mathrm{p}\left(\mathrm{x}\right)\:\mathrm{such}\:\mathrm{that} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\mathrm{p}\left(\mathrm{x}\right)\equiv\mathrm{8}\left(\mathrm{mod}\:\mathrm{x}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\mathrm{p}\left(\mathrm{x}\right)\equiv−\mathrm{24}\left(\mathrm{mod}\:\mathrm{x}+\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\mathrm{p}\left(\mathrm{x}\right)\equiv\mathrm{6}\left(\mathrm{mod}\:\mathrm{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\mathrm{p}\left(\mathrm{x}\right)\equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{x}+\mathrm{2}\right) \\ $$$$ \\…
Question Number 2160 by Yozzis last updated on 05/Nov/15 $${Solve}\:{the}\:{d}.{e} \\ $$$${sin}\left(\frac{{d}^{\mathrm{3}} {y}}{{dt}^{\mathrm{3}} }\right)+\mathrm{3}{t}^{\mathrm{2}} {y}=\mathrm{6}{t}. \\ $$ Commented by Yozzi last updated on 07/Nov/15 $${I}\:{was}\:{helping}\:{a}\:{friend}\:{with}\:{his}\:…
Question Number 2159 by Yozzis last updated on 05/Nov/15 $${Suppose}\:{that}\:{y}_{{n}\:} \:{satisfies}\:{the}\:{equations}\: \\ $$$$\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\frac{{d}^{\mathrm{2}} {y}_{{n}} }{{dx}^{\mathrm{2}} }−{x}\frac{{dy}_{{n}} }{{dx}}+{n}^{\mathrm{2}} {y}=\mathrm{0},\:{y}_{{n}} \left(\mathrm{1}\right)=\mathrm{1} \\ $$$${y}_{{n}} \left({x}\right)=\left(−\mathrm{1}\right)^{{n}} {y}_{{n}} \left(−{x}\right).…
Question Number 133228 by mnjuly1970 last updated on 20/Feb/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….{advanced}\:\:\:\:{calculus}…. \\ $$$$\:\:\:{prove}\:\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\psi\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}^{{n}} .{n}!}=−\sqrt{\mathrm{2}\pi}\:\left(\gamma+{ln}\left(\mathrm{2}\right)\right)…. \\ $$$$ \\ $$ Answered by Dwaipayan Shikari…
Question Number 2158 by Yozzis last updated on 05/Nov/15 $${Find}\:{the}\:{ratio},\:{over}\:{one}\:{revolution},\:{of}\:{the}\:{distance}\:{moved}\:{by} \\ $$$${a}\:{wheel}\:{rolling}\:{on}\:{a}\:{flat}\:{surface}\:{to}\:{the}\:{distance}\:{traced}\:{out}\:{by} \\ $$$${a}\:{point}\:{on}\:{its}\:{circumference}.\: \\ $$ Commented by ssahoo last updated on 06/Nov/15 $$\mathrm{Wheel}\:\mathrm{distance}=\mathrm{2}\pi{r} \\…
Question Number 133231 by SOMEDAVONG last updated on 20/Feb/21 $$\mathrm{Give}\:\mathrm{S}_{\mathrm{n}} =\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}^{\mathrm{n}+\mathrm{1}} }\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{2}^{\mathrm{k}} }{\mathrm{k}}\:\:.\mathrm{Find}\:\underset{\mathrm{n}\rightarrow+\propto} {\mathrm{lim}S}_{\mathrm{n}} \:. \\ $$ Answered by Ar Brandon last updated…
Question Number 67692 by mr W last updated on 30/Aug/19 Answered by mr W last updated on 30/Aug/19 Commented by mr W last updated on…
Question Number 2157 by Yozzi last updated on 05/Nov/15 $${Evaluate}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}+\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{1}!}+\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{2}!}+\frac{\mathrm{4}^{\mathrm{3}} }{\mathrm{3}!}+… \\ $$$${by}\:{considering}\:{the}\:{series}\:{expansion} \\ $$$${of}\:{an}\:{expression}\:{of}\:{the}\:{form}\:{P}\left({x}\right){e}^{{x}} \\ $$$${where}\:{P}\left({x}\right)\:{is}\:{a}\:{suitably}\:{chosen} \\ $$$${polynomial}\:{in}\:{x}.\: \\ $$$$…
Question Number 67688 by Rio Michael last updated on 30/Aug/19 $${A}\:{relation}\:\mathbb{R}\:{defined}\:{by}\:\:\:_{\left({x},{y}\right)} {R}_{\left({u},{v}\right)} \:\Leftrightarrow\:\:{v}^{\mathrm{2}} −{y}^{\mathrm{2}} \:=\:{u}^{\mathrm{2}} −{x}^{\mathrm{2}} \\ $$$${show}\:{that}\:{R}\:{is}\:{an}\:{equivalent}\:{Relation}. \\ $$ Commented by Prithwish sen last…