Question Number 26363 by d.monhbayr@gmail.com last updated on 24/Dec/17 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{1}+{x}\mathrm{sin}\:{x}}−\sqrt{\mathrm{cos}\:\mathrm{2}{x}}}{{tg}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}} \\ $$ Commented by kaivan.ahmadi last updated on 24/Dec/17 $$=\mathrm{li}\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{m}}\frac{\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }−\sqrt{\mathrm{1}−\mathrm{2x}^{\mathrm{2}} }}{\frac{\mathrm{x}^{\mathrm{2}}…
Question Number 91848 by jagoll last updated on 03/May/20 $${y}^{''} −\mathrm{4}{y}^{'} +\mathrm{4}{y}=\left({x}+\mathrm{1}\right){e}^{\mathrm{2}{x}} \\ $$$$ \\ $$ Commented by john santu last updated on 03/May/20 $${complementary}\:{solution}…
Question Number 26299 by d.monhbayr@gmail.com last updated on 23/Dec/17 $${y}=\frac{\mathrm{1}+{e}^{{x}} }{\mathrm{1}−{e}^{{x}} } \\ $$$${y}'=? \\ $$ Commented by abdo imad last updated on 24/Dec/17 $$\frac{{dy}}{{dx}}=\frac{{d}\left(\mathrm{1}+{e}^{{x}}…
Question Number 26300 by d.monhbayr@gmail.com last updated on 23/Dec/17 $${y}={a}^{\mathrm{arc}{tg}\sqrt{{x}}} \\ $$$${y}'=? \\ $$ Commented by abdo imad last updated on 24/Dec/17 $${we}\:{have}\:\:{y}=\:{e}^{{arctan}\left(\sqrt{\left.{x}\right)}{lna}\:\right.} \:\:\:{with}\:{condition}\:{a}>\mathrm{0} \\…
Question Number 26297 by d.monhbayr@gmail.com last updated on 23/Dec/17 $${y}={x}−\frac{\mathrm{2}}{{x}^{\mathrm{4}} }−\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{3}} } \\ $$$${y}'=? \\ $$ Answered by Joel578 last updated on 24/Dec/17 $$\:{y}\:=\:{x}\:−\:\mathrm{2}{x}^{−\mathrm{4}} \:−\:\frac{\mathrm{1}}{\mathrm{3}}{x}^{−\mathrm{3}}…
Question Number 91825 by john santu last updated on 03/May/20 $$\left(\mathrm{cos}\:{x}\right)\:\frac{{dy}}{{dx}}−{y}\left(\mathrm{sin}\:{x}\right)\:=\:\mathrm{cot}\:\left({x}\right) \\ $$ Commented by Tony Lin last updated on 03/May/20 $$\left({ycosx}\right)'={cotx} \\ $$$${ycosx}=\int{cotxdx}={ln}\mid{sinx}\mid+{c} \\…
Question Number 26240 by sorour87 last updated on 22/Dec/17 $${y}^{\left(\mathrm{2}\right)} −{y}=\left(\mathrm{1}−{e}^{\mathrm{2}{x}} \right)^{\frac{−\mathrm{1}}{\mathrm{2}}} \\ $$ Commented by prakash jain last updated on 23/Dec/17 $${y}^{\left(\mathrm{2}\right)} −{y}=\left(\mathrm{1}−{e}^{\mathrm{2}{x}} \right)^{\frac{−\mathrm{1}}{\mathrm{2}}}…
Question Number 91770 by john santu last updated on 03/May/20 $${y}''+\mathrm{3}{y}={x}^{\mathrm{3}} +\mathrm{3}{x} \\ $$ Commented by Joel578 last updated on 03/May/20 $$\mathrm{Let}\:\mathrm{the}\:\mathrm{particular}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{in}\:\mathrm{form} \\ $$$${y}_{{p}} \:=\:{Ax}^{\mathrm{3}}…
Question Number 91750 by jagoll last updated on 02/May/20 $$\left(\mathrm{2}{x}−{y}+\mathrm{1}\right){dx}\:=\:\left({x}−\mathrm{4}{y}+\mathrm{3}\right){dy} \\ $$ Commented by mr W last updated on 02/May/20 $${see}\:{Q}\mathrm{90770} \\ $$ Commented by…
Question Number 26198 by sorour87 last updated on 22/Dec/17 $$\left(\mathrm{2}{x}\sqrt{{x}}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right){dx}+\mathrm{2}{y}\sqrt{{x}}{dy}=\mathrm{0} \\ $$ Answered by ajfour last updated on 22/Dec/17 $${let}\:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \\…