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Category: Differential Equation

Find-the-sum-r-0-n-3r-5-n-r-5-n-0-8-n-1-11-n-2-3n-5-n-n-as-a-simple-function-of-n-

Question Number 519 by Yugi last updated on 25/Jan/15 $${Find}\:{the}\:{sum}\:\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\left(\mathrm{3}{r}+\mathrm{5}\right)\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}=\mathrm{5}\begin{pmatrix}{{n}}\\{\mathrm{0}}\end{pmatrix}\:+\mathrm{8}\begin{pmatrix}{{n}}\\{\mathrm{1}}\end{pmatrix}\:+\mathrm{11}\begin{pmatrix}{{n}}\\{\mathrm{2}}\end{pmatrix}\:+…\left(\mathrm{3}{n}+\mathrm{5}\right)\begin{pmatrix}{{n}}\\{{n}}\end{pmatrix}\: \\ $$$${as}\:{a}\:{simple}\:{function}\:{of}\:{n}. \\ $$ Commented by prakash jain last updated on 22/Jan/15 $$\underset{{r}=\mathrm{0}}…

A-person-is-said-to-be-n-years-old-where-n-is-a-non-negative-integer-if-the-person-has-lived-at-least-n-years-and-has-not-lived-n-1-years-At-some-point-Tom-is-4-years-old-and-John-is-three-times-

Question Number 517 by Yugi last updated on 25/Jan/15 $${A}\:{person}\:{is}\:{said}\:{to}\:{be}\:{n}\:{years}\:{old}\:\left(\:{where}\:{n}\:{is}\:{a}\:{non}−{negative}\:{integer}\right)\:{if}\: \\ $$$${the}\:{person}\:{has}\:{lived}\:{at}\:{least}\:{n}\:{years}\:{and}\:{has}\:{not}\:{lived}\:{n}+\mathrm{1}\:{years}.\:{At}\:{some}\:{point} \\ $$$${Tom}\:{is}\:\mathrm{4}\:{years}\:{old}\:{and}\:{John}\:{is}\:{three}\:{times}\:{as}\:{old}\:{as}\:{Mary}.\:{At}\:{another}\:{time}, \\ $$$${Mary}\:{is}\:{twice}\:{as}\:{old}\:{as}\:{Tom}\:{and}\:{John}\:{is}\:{five}\:{times}\:{as}\:{old}\:{as}\:{Tom}.\:{At}\:{a}\:{third}\: \\ $$$${time},\:{John}\:{is}\:{twice}\:{as}\:{old}\:{as}\:{Mary}\:{and}\:{Tom}\:{is}\:{t}\:{years}\:{old}.\:{What}\:{is}\:{the}\:{largest} \\ $$$${possible}\:{value}\:{of}\:{t}? \\ $$ Commented by prakash…

Use-Euler-method-to-estimate-the-value-of-solution-at-given-point-x-dy-dx-5xe-x-5-y-0-1-dx-0-1-and-x-1-

Question Number 131551 by liberty last updated on 06/Feb/21 $$\mathrm{Use}\:\mathrm{Euler}\:\mathrm{method}\:\mathrm{to}\:\mathrm{estimate}\: \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{solution}\:\mathrm{at}\:\mathrm{given}\: \\ $$$$\mathrm{point}\:\mathrm{x}^{\ast} \:.\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{5xe}^{\mathrm{x}^{\mathrm{5}} } \:,\mathrm{y}\left(\mathrm{0}\right)=\mathrm{1}\:,\mathrm{dx}=\mathrm{0}.\mathrm{1} \\ $$$$\mathrm{and}\:\mathrm{x}^{\ast} =\mathrm{1}. \\ $$ Terms of Service…