Question Number 103014 by bobhans last updated on 12/Jul/20 $${y}''\:−{y}\:=\:\frac{{e}^{{x}} }{{e}^{{x}} \:+\:{e}^{−{x}} }\:. \\ $$ Answered by bobhans last updated on 12/Jul/20 $${the}\:{characteristic}\:{equation}\:{of}\:{the} \\ $$$${homogenous}\:{equation}\:{is}\:\gamma^{\mathrm{2}}…
Question Number 103009 by bobhans last updated on 12/Jul/20 $$\left({D}^{\mathrm{2}} −\mathrm{4}{D}+\mathrm{4}\right){y}\:=\:{x}^{\mathrm{3}} {e}^{\mathrm{2}{x}} \: \\ $$ Answered by bramlex last updated on 12/Jul/20 $$\mathrm{since}\:\mathrm{y}\:=\:\mathrm{e}^{\mathrm{2x}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{part}\:\mathrm{of}\:\mathrm{the} \\…
Question Number 102840 by Dwaipayan Shikari last updated on 11/Jul/20 $${y}'+\sqrt{{x}+{y}−\mathrm{1}}={x}+{y}+\mathrm{1} \\ $$ Answered by OlafThorendsen last updated on 12/Jul/20 $${y}\:=\:\mathrm{Y}^{\mathrm{2}} −{x}+\mathrm{1} \\ $$$$\mathrm{2Y}'\mathrm{Y}−\mathrm{1}+\sqrt{\mathrm{Y}^{\mathrm{2}} }\:=\:\mathrm{Y}^{\mathrm{2}}…
Question Number 168325 by cortano1 last updated on 08/Apr/22 $$\:\:\:\:\:\:\mathrm{2}{yy}''−\left({y}'\right)^{\mathrm{2}} \:−\mathrm{1}=\:\frac{\mathrm{8}{y}^{\mathrm{2}} }{\mathrm{cos}\:^{\mathrm{2}} {x}}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 168276 by Florian last updated on 07/Apr/22 $$\:\:\:\:{y}\:=\:\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}} \\ $$$$\:\:\:\:{y}'\:=\: \\ $$$$\:\:\:\:\: \\ $$ Commented by cortano1 last updated on 07/Apr/22 $$\:{y}'=\frac{\mathrm{1}+\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}}{\mathrm{2}\sqrt{{x}+\sqrt{{x}}}}}{\mathrm{2}\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}} \\…
Question Number 168267 by Florian last updated on 07/Apr/22 $$\:\:\:\:\:\:{Solve}\:\:{this}\:{integral}\:: \\ $$$$\:\:\:\:\:\:\:\int\frac{\sqrt{{x}+\mathrm{1}}−\mathrm{1}}{\:\sqrt{{x}+\mathrm{1}}+\mathrm{1}} \\ $$$$ \\ $$ Answered by MJS_new last updated on 07/Apr/22 $$\int\frac{\sqrt{{x}+\mathrm{1}}−\mathrm{1}}{\:\sqrt{{x}+\mathrm{1}}+\mathrm{1}}{dx}= \\…
Question Number 102716 by Ar Brandon last updated on 10/Jul/20 $$\mathrm{y}''+\mathrm{3y}'−\mathrm{10y}=\mathrm{14e}^{−\mathrm{5x}} \\ $$ Answered by bramlex last updated on 10/Jul/20 $$\mathrm{Homogenous}\: \\ $$$$\mathrm{r}^{\mathrm{2}} +\mathrm{3r}−\mathrm{10}=\mathrm{0}\:\Rightarrow\left(\mathrm{r}+\mathrm{5}\right)\left(\mathrm{r}−\mathrm{2}\right)=\mathrm{0} \\…
Question Number 102639 by bobhans last updated on 10/Jul/20 $$\frac{{dy}}{{dx}}\:−\mathrm{2}{xy}\:=\:\mathrm{6}{y}\:{e}^{{y}^{\mathrm{2}} } \\ $$ Answered by Ar Brandon last updated on 10/Jul/20 $$\mathrm{Let}\:\mathrm{y}=\mathrm{vx}\:\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}} \\ $$$$\Rightarrow\:\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}−\mathrm{2vx}^{\mathrm{2}} =\mathrm{6vxe}^{\mathrm{v}^{\mathrm{2}}…
Question Number 102527 by bemath last updated on 09/Jul/20 $$\mathrm{2}{y}''−{y}'+{y}\:=\:\mathrm{cos}\:\mathrm{3}{x}\: \\ $$ Answered by Ar Brandon last updated on 09/Jul/20 $$\mathrm{Let}\:\mathrm{y}=\mathrm{acos3x}+\mathrm{bsin3x} \\ $$$$\Rightarrow\mathrm{y}'=−\mathrm{3asin3x}+\mathrm{3bcos3x}\:,\:\mathrm{y}''=−\mathrm{9acos3x}−\mathrm{9bsin3x} \\ $$$$\Rightarrow−\mathrm{2}×\mathrm{9}\left(\mathrm{acos3x}+\mathrm{bsin3x}\right)−\mathrm{3}\left(\mathrm{bcos3x}−\mathrm{asin3x}\right)+\mathrm{acos3x}+\mathrm{bsin3x}…
Question Number 102515 by bemath last updated on 09/Jul/20 $$\left(\frac{{x}}{{y}}\right){y}'=\:\frac{\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}}{{x}+\mathrm{1}} \\ $$ Answered by bobhans last updated on 09/Jul/20 $$\frac{{dy}}{{y}\left(\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}\right)}\:=\:\frac{{dx}}{{x}\left({x}+\mathrm{1}\right)} \\ $$$$\left(\mathrm{1}\right)\:\frac{\mathrm{1}}{{y}\left(\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}\right)}\:=\:\frac{{A}}{{y}}\:+\:\frac{{By}+{C}}{\mathrm{2}{y}^{\mathrm{2}}…