Question Number 161039 by HongKing last updated on 11/Dec/21 $$\mathrm{let}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equation}: \\ $$$$\left(\mathrm{1}\:+\:\mathrm{x}\right)\:\mathrm{y}^{''} \left(\mathrm{x}\right)\:+\:\left(\mathrm{1}\:-\:\mathrm{x}\right)\:\mathrm{y}^{'} \left(\mathrm{x}\right)\:=\:\frac{\mathrm{1}-\mathrm{x}}{\mathrm{1}+\mathrm{x}}\:\mathrm{y}\left(\mathrm{x}\right) \\ $$$$\mathrm{y}\left(\mathrm{0}\right)\:=\:\mathrm{1}\:,\:\mathrm{y}^{'} \left(\mathrm{0}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\left(\mathrm{y}^{''} \left(\mathrm{x}\right)\:+\:\mathrm{y}^{'} \left(\mathrm{x}\right)\:+\:\mathrm{y}\left(\mathrm{x}\right)\right)\:\mathrm{e}^{-\boldsymbol{\mathrm{x}}}…
Question Number 161035 by cortano last updated on 11/Dec/21 $${Solve}\:{for}\:{x}\:\epsilon\:\mathbb{R}\: \\ $$$$\:\sqrt{\sqrt{\mathrm{3}}−{x}}\:=\:{x}\sqrt{\sqrt{\mathrm{3}}+{x}}\: \\ $$ Answered by Rasheed.Sindhi last updated on 11/Dec/21 $$\:\sqrt{\sqrt{\mathrm{3}}−{x}}\:=\:{x}\sqrt{\sqrt{\mathrm{3}}+{x}}\: \\ $$$$\sqrt{\mathrm{3}}\:−{x}={x}^{\mathrm{2}} \left(\sqrt{\mathrm{3}}\:+{x}\right)…
Question Number 161032 by mathlove last updated on 11/Dec/21 Answered by cortano last updated on 11/Dec/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 161023 by cortano last updated on 10/Dec/21 Commented by som(math1967) last updated on 11/Dec/21 $$\mathrm{100}\:{squnits}\:? \\ $$ Commented by cortano last updated on…
Question Number 161010 by HongKing last updated on 10/Dec/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 161011 by HongKing last updated on 10/Dec/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 161000 by bobhans last updated on 10/Dec/21 $$\:\:\sqrt{\mathrm{1}−\frac{\mathrm{x}+\mathrm{1}}{\mathrm{x}}}\:+\:\mid\mathrm{x}−\mathrm{3}\mid\:\geqslant\:\mathrm{0}\: \\ $$ Commented by cortano last updated on 10/Dec/21 $$\:\:\:\mathrm{x}<\mathrm{0} \\ $$ Terms of Service…
Question Number 95469 by 6532 last updated on 25/May/20 $$\left(\mathrm{9b}^{\mathrm{2}} −\mathrm{25}\right) \\ $$$$\mathrm{why}\:\mathrm{is}\:\mathrm{this}\:\mathrm{inside}\:\mathrm{the}\:\mathrm{bracket}\:\mathrm{as}\:\mathrm{it}\:\mathrm{is}\:\mathrm{a}\:\mathrm{diffetence}\:\mathrm{of}\:\mathrm{two}\:\mathrm{squares}? \\ $$ Commented by PRITHWISH SEN 2 last updated on 25/May/20 $$\left(\mathrm{3b}−\mathrm{5}\right)\left(\mathrm{3b}+\mathrm{5}\right)…
Question Number 160998 by Tawa11 last updated on 10/Dec/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 95447 by Tony Lin last updated on 25/May/20 $${can}\:{I}\:{write}\:{the}\:{solution}\:{of} \\ $$$${ay}''+{by}'+{cy}=\mathrm{0} \\ $$$${y}=\begin{cases}{{c}_{\mathrm{1}} {e}^{\frac{−{b}+\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}}{x}} +{c}_{\mathrm{2}} {e}^{\frac{−{b}−\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}}{x}} ,{when}\:{b}^{\mathrm{2}} −\mathrm{4}{ac}\neq\mathrm{0}}\\{{c}_{\mathrm{1}} {e}^{\frac{−{b}}{\mathrm{2}}{x}} +{c}_{\mathrm{2}} {xe}^{\frac{−{b}}{\mathrm{2}}{x}}…