Question Number 61181 by mathsolverby Abdo last updated on 30/May/19 $${sove}\:\left(\mathrm{1}+{e}^{−{x}} \right){y}^{''} \:+\left(\mathrm{2}+{e}^{{x}} \right){y}^{'} \:=\left({x}+\mathrm{1}\right){e}^{{x}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 61180 by mathsolverby Abdo last updated on 30/May/19 $${solve}\:{y}^{''} \:+\mathrm{3}{y}^{'} −{y}\:={sin}\left(\mathrm{2}{x}\right) \\ $$ Answered by tanmay last updated on 30/May/19 $$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\mathrm{3}\frac{{dy}}{{dx}}−{y}={sin}\mathrm{2}{x}…
Question Number 61178 by mathsolverby Abdo last updated on 30/May/19 $${solve}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}^{'} \:+\left(\mathrm{1}−{x}^{\mathrm{2}} \right){y}\:={x}\:{e}^{−\mathrm{3}{x}} \\ $$ Commented by maxmathsup by imad last updated on 31/May/19…
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Question Number 126641 by TITA last updated on 22/Dec/20 Commented by TITA last updated on 22/Dec/20 $$\mathrm{please}\:\mathrm{help} \\ $$ Answered by Lordose last updated on…
Question Number 126635 by BHOOPENDRA last updated on 22/Dec/20 Answered by Ar Brandon last updated on 22/Dec/20 $$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} =\mathrm{r}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \theta\mathrm{cos}^{\mathrm{2}} \phi+\mathrm{r}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}}…
Question Number 126631 by mnjuly1970 last updated on 23/Dec/20 $$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…{calculus}… \\ $$$$\:\:\:\:\:{prove}\:{that}:: \\ $$$$\:\:\:\:::\:\:\:\:\:\:\Omega\overset{??} {=}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{n}−\mathrm{1}} {ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right){dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{2}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{{H}_{{k}}…
Question Number 126605 by bramlexs22 last updated on 22/Dec/20 $$\:{Let}\:{P}\:{be}\:{point}\:{on}\:{the}\:{graph}\: \\ $$$${of}\:{a}\:{straight}\:{line}\:{y}=\mathrm{2}{x}−\mathrm{3}\:{and}\:{Q} \\ $$$${be}\:{a}\:{point}\:{on}\:{the}\:{graph}\:{of}\:{a}\:{parabola} \\ $$$${y}={x}^{\mathrm{2}} +{x}+\mathrm{1}\:.{Find}\:{the}\:{shortest}\: \\ $$$${distance}\:{between}\:{P}\:{and}\:{Q}\:. \\ $$ Answered by liberty last…
Question Number 126524 by rs4089 last updated on 21/Dec/20 Answered by Olaf last updated on 21/Dec/20 $$\mathrm{Let}\:{u}\:=\:{e}^{{x}} \\ $$$$\frac{{dy}}{{dx}}\:=\:\frac{{dy}}{{du}}.\frac{{du}}{{dx}}\:=\:{e}^{{x}} \frac{{dy}}{{du}}\:=\:{u}\frac{{dy}}{{du}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:=\:\frac{{d}}{{dx}}\left(\frac{{du}}{{dx}}\right)\:=\:\frac{{d}}{{du}}\left({u}\frac{{du}}{{du}}\right)\frac{{du}}{{dx}} \\…
Question Number 192054 by cortano12 last updated on 07/May/23 $$\:\:\:\mathrm{If}\:\mathrm{Q}\:=\:\frac{\mathrm{2}−\mathrm{x}}{\mathrm{y}−\mathrm{1}}\:;\:−\mathrm{5}\leqslant\mathrm{x}<−\mathrm{1}\:,\:\mathrm{5}\leqslant\mathrm{y}<\mathrm{6} \\ $$$$\:\:\:\mathrm{Find}\:\mathrm{Q}_{\mathrm{max}} .\: \\ $$ Answered by mehdee42 last updated on 07/May/23 $$−\mathrm{5}\leqslant{x}<−\mathrm{1}\overset{×−\mathrm{1}} {\Rightarrow}\:\mathrm{1}<−{x}\leqslant\mathrm{5}\overset{+\mathrm{2}} {\Rightarrow}\mathrm{3}<\mathrm{2}−{x}\leqslant\mathrm{7}\:\:\left({i}\right)…