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solve-y-y-tanx-

Question Number 111011 by mohammad17 last updated on 01/Sep/20 $${solve}:\:{y}^{''} +{y}^{'} ={tanx} \\ $$ Answered by mathmax by abdo last updated on 01/Sep/20 $$\left.\mathrm{h}\right)\rightarrow\mathrm{r}^{\mathrm{2}} \:+\mathrm{r}\:=\mathrm{0}\:\Rightarrow\mathrm{r}\left(\mathrm{r}+\mathrm{1}\right)\:=\mathrm{0}\Rightarrow\mathrm{r}=\mathrm{0}\:\mathrm{or}\:\mathrm{r}=−\mathrm{1}\:\Rightarrow\mathrm{y}\left(\mathrm{x}\right)=\mathrm{ae}^{−\mathrm{x}}…

Question-176540

Question Number 176540 by yaslm last updated on 20/Sep/22 Answered by Peace last updated on 21/Sep/22 $$\mathcal{F}\left({x}\right)=\int\frac{{dx}}{{cos}^{\mathrm{2}} \left({x}\right)\sqrt{{cos}\left({x}\right)}}={tg}\left({x}\right)\frac{\mathrm{1}}{\:\sqrt{{cos}\left({x}\right)}}−\int{tg}\left({x}\right).\frac{{sin}\left({x}\right)}{\mathrm{2}\sqrt{{cos}\left({x}\right)}{cos}\left({x}\right)} \\ $$$$−\int\frac{{sin}^{\mathrm{2}} \left({x}\right)}{\mathrm{2}{cos}^{\mathrm{2}} \left({x}\right)\sqrt{{cos}\left({x}\right)}}{dx} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}\mathcal{F}\left({x}\right)=\frac{{sin}\left({x}\right)}{{cos}\left({x}\right)\sqrt{{cos}\left({x}\right)}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\:\sqrt{{cos}\left({x}\right)}} \\…

find-the-of-differnt-ways-in-which-a-cricket-team-consisting-of-11people-can-be-choosen-from-the-group-of-16-

Question Number 45433 by mondodotto@gmail.com last updated on 12/Oct/18 $$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{differnt}}\:\boldsymbol{\mathrm{ways}}\: \\ $$$$\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{which}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{cricket}}\:\boldsymbol{\mathrm{team}}\:\boldsymbol{\mathrm{consisting}}\:\boldsymbol{\mathrm{of}} \\ $$$$\mathrm{11}\boldsymbol{\mathrm{people}},\boldsymbol{\mathrm{can}}\:\boldsymbol{\mathrm{be}}\:\boldsymbol{\mathrm{choosen}} \\ $$$$\boldsymbol{\mathrm{from}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{group}}\:\boldsymbol{\mathrm{of}}\:\mathrm{16} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 13/Oct/18…