Question Number 119464 by Engr_Jidda last updated on 24/Oct/20 $$\mathrm{Expand}\:\mathrm{as}\:\mathrm{far}\:\mathrm{as}\:\mathrm{the}\:\mathrm{term}\:\mathrm{in}\:\mathrm{x}^{\mathrm{3}} \\ $$$$\left(\mathrm{1}\right)\:\left(\mathrm{3}−\mathrm{2x}−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{9}} \\ $$$$\left(\mathrm{2}\right)\:\left(\mathrm{1}−\mathrm{x}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{8}} \\ $$ Answered by TANMAY PANACEA last updated on…
Question Number 184937 by MikeH last updated on 14/Jan/23 $$\mathrm{find}\:\mathrm{the}\:\mathrm{laplace}\:\mathrm{transform}\:\mathrm{of}\:\mathrm{the}\:\mathrm{differential} \\ $$$$\mathrm{equation}\:\mathrm{below} \\ $$$$\frac{{dy}}{{dt}}\:+\:\mathrm{5}{y}\left({t}\right)\:+\:\mathrm{6}\int_{\mathrm{0}} ^{{t}} {y}\left(\tau\right){d}\tau\:=\:{u}\left({t}\right)\:\mathrm{where}\:{y}\left(\mathrm{0}\right)\:=\:\mathrm{2} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 184936 by MikeH last updated on 14/Jan/23 $$\mathrm{Use}\:\mathrm{Laplace}\:\mathrm{transform}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{differential} \\ $$$$\mathrm{equation} \\ $$$$\:\frac{{d}^{\mathrm{2}} {v}\left({t}\right)}{{dt}^{\mathrm{2}} }\:+\mathrm{6}\frac{{dv}\left({t}\right)}{{dt}}\:+\:\mathrm{8}{v}\left({t}\right)\:=\:\mathrm{2}{u}\left({t}\right)\:\: \\ $$$$\mathrm{when}\:{v}\left(\mathrm{0}\right)\:=\:\mathrm{1}\:\mathrm{and}\:\overset{\bullet} {{v}}\left(\mathrm{0}\right)\:=\:−\mathrm{2} \\ $$ Answered by hmr last…
Question Number 119290 by bobhans last updated on 23/Oct/20 $$\:\:\left(\mathrm{3}{x}−{y}+\mathrm{1}\right)\:{dx}\:+\left(\mathrm{6}{x}+\mathrm{2}{y}−\mathrm{3}\right)\:{dy}\:=\:\mathrm{0}\: \\ $$ Answered by TANMAY PANACEA last updated on 23/Oct/20 $${dy}\left(\mathrm{6}{x}+\mathrm{2}{y}−\mathrm{3}\right)=\left(−\mathrm{3}{x}+{y}−\mathrm{1}\right){dx} \\ $$$$\frac{{dy}}{{dx}}=\frac{−\mathrm{3}{x}+{y}−\mathrm{1}}{\mathrm{6}{x}+\mathrm{2}{y}−\mathrm{3}} \\ $$$${x}={X}+{h}…
Question Number 119235 by bemath last updated on 23/Oct/20 $$\:\left({D}^{\mathrm{3}} +{D}^{\mathrm{2}} −\mathrm{4}{D}−\mathrm{4}\right){y}\:=\:{e}^{\mathrm{4}{x}} \\ $$ Answered by benjo_mathlover last updated on 23/Oct/20 $$\left({D}+{I}\right)\left({D}−\mathrm{2}{I}\right)\left({D}+\mathrm{2}{I}\right)\left({y}\right)\:=\:{e}^{\mathrm{4}{x}} \\ $$$${let}\:\left({D}−\mathrm{2}{I}\right)\left({D}+\mathrm{2}{I}\right)\left({y}\right)\:=\:{z} \\…
Question Number 119233 by bemath last updated on 23/Oct/20 $$\:\left({D}^{\mathrm{2}} −\mathrm{5}{D}+\mathrm{6}\right){y}\:=\:{e}^{\mathrm{3}{x}} \\ $$ Answered by benjo_mathlover last updated on 23/Oct/20 $$\:\left({D}−\mathrm{2}{I}\right)\left({D}−\mathrm{3}{I}\right)\left({y}\right)\:=\:{e}^{\mathrm{3}{x}} \:;\:{I}\left({y}\right)={y} \\ $$$${let}\:\left({D}−\mathrm{3}{I}\right)\left({y}\right)\:=\:{z} \\…
Question Number 119065 by benjo_mathlover last updated on 21/Oct/20 $${solve}\:\left({D}^{\mathrm{2}} −\mathrm{2}{D}+\mathrm{1}\right){y}\:=\:{e}^{{x}} \:\mathrm{ln}\:{x}\: \\ $$$${by}\:{using}\:{the}\:{method}\:{of}\:{variation} \\ $$$${of}\:{parameters}. \\ $$ Answered by Olaf last updated on 22/Oct/20…
Question Number 119002 by benjo_mathlover last updated on 21/Oct/20 $$\frac{{dy}}{{dx}}\:=\:\frac{{xy}}{{x}^{\mathrm{2}} +\mathrm{1}}\:{and}\:{y}\left(\sqrt{\mathrm{15}}\right)\:=\:\mathrm{2} \\ $$ Answered by bramlexs22 last updated on 21/Oct/20 $$\:\frac{{dy}}{{y}}\:=\:\frac{{x}\:{dx}}{{x}^{\mathrm{2}} +\mathrm{1}}\:\Rightarrow\:\int\:\frac{{dy}}{{y}}\:=\:\int\:\frac{{x}\:{dx}}{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{ln}\:\left({y}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({x}^{\mathrm{2}}…
Question Number 118975 by benjo_mathlover last updated on 21/Oct/20 $$\:\left(\mathrm{3}{x}−\mathrm{5}{y}\right)\:{dx}\:+\:\left({x}+{y}\right)\:{dy}\:=\:\mathrm{0} \\ $$ Answered by 1549442205PVT last updated on 21/Oct/20 $$\:\left(\mathrm{3}{x}−\mathrm{5}{y}\right)\:{dx}\:+\:\left({x}+{y}\right)\:{dy}\:=\:\mathrm{0}\left(\mathrm{1}\right) \\ $$$$\mathrm{Put}\:\mathrm{y}=\mathrm{xt}\Rightarrow\mathrm{dy}=\mathrm{xdt}+\mathrm{tdx} \\ $$$$\left(\mathrm{1}\right)\Leftrightarrow\left(\mathrm{3x}−\mathrm{5xt}\right)\mathrm{dx}+\left(\mathrm{x}+\mathrm{xt}\right)\left(\mathrm{xdt}+\mathrm{tdx}\right) \\…
Question Number 118962 by benjo_mathlover last updated on 21/Oct/20 $$\:{x}\:{dy}\:=\:\left({y}+\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:\right)\:{dx}\:;\:{y}\left(\sqrt{\mathrm{3}}\right)\:=\:\mathrm{1} \\ $$ Answered by TANMAY PANACEA last updated on 21/Oct/20 $$\frac{{xdy}−{ydx}}{{x}^{\mathrm{2}} }=\frac{\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}}…