Question Number 103203 by bobhans last updated on 13/Jul/20 $${y}''+\mathrm{4}{y}'+\mathrm{4}{y}\:=\:\frac{{e}^{−\mathrm{2}{x}} }{{x}^{\mathrm{2}} }\: \\ $$ Answered by bramlex last updated on 14/Jul/20 $$\mathrm{characteristic}\:\mathrm{eq}\: \\ $$$$\chi^{\mathrm{2}} +\mathrm{4}\chi+\mathrm{4}\:=\:\mathrm{0}\:;\:\left(\chi+\mathrm{2}\right)^{\mathrm{2}}…
Question Number 103193 by bemath last updated on 13/Jul/20 $$\frac{{dy}}{{dx}}\:+\:{y}.\mathrm{cot}\:\left({x}\right)\:=\:\mathrm{sin}\:\left({x}\right) \\ $$ Commented by bobhans last updated on 13/Jul/20 Terms of Service Privacy Policy Contact:…
Question Number 103138 by bemath last updated on 13/Jul/20 $$\left({D}^{\mathrm{2}} −\mathrm{4}{D}\right){y}\:=\:{x}^{\mathrm{2}} \:{e}^{\mathrm{2}{x}} \\ $$ Commented by bemath last updated on 14/Jul/20 Commented by Israrchajk724 last…
Question Number 168663 by DomaPeti last updated on 15/Apr/22 $${how}\:{to}\:{solve}? \\ $$$$\frac{{cos}\left(\mathrm{30}°\right)\centerdot{sin}\left(\mathrm{45}°\right)−{cos}\left(\mathrm{30}°+{dx}\right)\centerdot{sin}\left(\mathrm{45}°+{dx}\right)}{{dx}}=? \\ $$$${dx}\rightarrow\mathrm{0} \\ $$ Commented by cortano1 last updated on 15/Apr/22 $$\underset{{dx}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\mathrm{6}}−\mathrm{cos}\:\left(\mathrm{30}°+{dx}\right).\mathrm{sin}\:\left(\mathrm{45}°+{dx}\right)}{{dx}}…
Question Number 103037 by bramlex last updated on 12/Jul/20 $$\mathrm{4x}^{\mathrm{2}} \mathrm{y}''\:+\mathrm{y}\:=\:\mathrm{0}\:,\:\mathrm{x}\:>\:\mathrm{0} \\ $$ Answered by maths mind last updated on 12/Jul/20 $${let}\:{y}={x}^{{t}} \\ $$$$\Rightarrow\mathrm{4}{x}^{\mathrm{2}} \left({t}\left({t}−\mathrm{1}\right)\right){x}^{{t}−\mathrm{2}}…
Question Number 103014 by bobhans last updated on 12/Jul/20 $${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}}}}} \\…