Question Number 82350 by jagoll last updated on 20/Feb/20 $$\mathrm{2}{y}''\:+\mathrm{5}{y}'\:−\mathrm{3}{y}=\mathrm{0} \\ $$$${general}\:{and}\:{particular}\:{solution} \\ $$$${when}\:{x}\:=\mathrm{0}\:,\:{y}\:=\:\mathrm{4}\:{and}\:\frac{{dy}}{{dx}}\:=\:\mathrm{9}? \\ $$ Commented by jagoll last updated on 20/Feb/20 Terms of…
Question Number 82333 by jagoll last updated on 20/Feb/20 $$\left({y}^{\mathrm{4}} −\mathrm{2}{xy}\right)\:{dx}\:=\:−\mathrm{3}{x}^{\mathrm{2}} \:{dy} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 16788 by jyoti5726 last updated on 26/Jun/17 Commented by 1234Hello last updated on 04/Jul/17 $$\mathrm{tan}^{−\mathrm{1}} \:\sqrt{\mathrm{3}}\:=\:\frac{\pi}{\mathrm{3}} \\ $$$$\mathrm{sec}^{−\mathrm{1}} \:\left(−\mathrm{2}\right)\:=\:\mathrm{cos}^{−\mathrm{1}} \:\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)\:=\:\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\mathrm{tan}^{−\mathrm{1}} \:\sqrt{\mathrm{3}}\:−\:\mathrm{sec}^{−\mathrm{1}}…
Question Number 82303 by Power last updated on 20/Feb/20 Commented by Power last updated on 20/Feb/20 $$\mathrm{thanks}\: \\ $$ Commented by Power last updated on…
Question Number 82247 by jagoll last updated on 19/Feb/20 $$\left({D}^{\mathrm{4}} +\mathrm{2}{D}^{\mathrm{2}} +\mathrm{1}\right){y}\:={x}^{\mathrm{2}} \:\mathrm{cos}\:{x}\: \\ $$ Commented by john santu last updated on 20/Feb/20 $${characteristic}\:{equation}\: \\…
Question Number 82245 by jagoll last updated on 19/Feb/20 $${find}\:{the}\:{solution}\: \\ $$$${x}\:\mathrm{sin}\:\left(\frac{{y}}{{x}}\right)\:{dy}\:=\:\left[{y}\:\mathrm{sin}\:\left(\frac{{y}}{{x}}\right)\:−{x}\right]\:{dx} \\ $$ Commented by john santu last updated on 19/Feb/20 $${let}\:{v}\:=\:\frac{{y}}{{x}}\:\Rightarrow\:{y}\:=\:{vx} \\ $$$$\frac{{dy}}{{dx}}\:=\:{v}\:+\:\frac{{dv}}{{dx}}\:\Rightarrow\:{dy}\:={v}\:{dx}+\:{dv}\:…
Question Number 16689 by Nayon last updated on 25/Jun/17 $${why}\:{any}\:{infinitely}\:{differentiable}\:{function}\:{is}\:{a}\:{power}\:{series}? \\ $$ Answered by prakash jain last updated on 25/Jun/17 $$\mathrm{Taylor}\:\mathrm{Series}\:\mathrm{for}\:\mathrm{class}\:\mathrm{C}^{\infty} . \\ $$ Terms…
Question Number 147753 by Odhiambojr last updated on 23/Jul/21 $${Find}\:{a}\:{point}\:{on}\:{the}\:{curve} \\ $$$$\:{y}={x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{3}\:{at}\:{which}\:{the}\:{tangent}\:{is} \\ $$$$\:{parallel}\:{to}\:{the}\:{x}-{axis} \\ $$ Answered by Olaf_Thorendsen last updated on 23/Jul/21 $$\frac{{dy}}{{dx}}\:=\:\mathrm{2}{x}−\mathrm{2}\:=\:\mathrm{2}\left({x}−\mathrm{1}\right)…
Question Number 147748 by Odhiambojr last updated on 23/Jul/21 $${prove}\:{that}\:{curves}\:{x}^{\mathrm{2}\:} −{y}^{\mathrm{2}} =\mathrm{3}\:{and} \\ $$$$\:{xy}=\mathrm{2}\:{intersect}\:{at}\:{the}\:{right}\:{angle}\: \\ $$$$ \\ $$ Commented by Olaf_Thorendsen last updated on 23/Jul/21…
Question Number 147713 by Odhiambojr last updated on 22/Jul/21 $${Find}\:{a}\:{point}\:{on}\:{the}\:{curve}\:{y}=\sqrt{{x}} \\ $$$${where}\:{the}\:{tangent}\:{makes}\:{an}\:{angle}\: \\ $$$$\mathrm{45}°\:{with}\:{the}\:{positive}\:{x}-{axis} \\ $$ Answered by Olaf_Thorendsen last updated on 22/Jul/21 $$\frac{{dy}}{{dx}}\:=\:\mathrm{tan}\left(\theta\left({x}\right)\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}} \\…