Question Number 99072 by Mr.D.N. last updated on 18/Jun/20 Answered by mr W last updated on 18/Jun/20 $$\int\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{2}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}}{dx} \\…
Question Number 98993 by Rio Michael last updated on 17/Jun/20 $$\mathrm{Use}\:\mathrm{the}\:\mathrm{laplace}\:\mathrm{tranform}\:\mathrm{to}\:\mathrm{solve}\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{5}\frac{{dy}}{{dx}}\:+\:\mathrm{6}{y}\:=\:{e}^{−{x}} \\ $$$$\mathrm{for}\:\:{y}\:=\:\mathrm{0},\:\mathrm{and}\:\frac{{dy}}{{dx}}\:=\:\mathrm{1}\:\mathrm{when}\:{x}\:=\:\mathrm{0} \\ $$ Answered by mathmax by abdo last updated on…
Question Number 98808 by bemath last updated on 16/Jun/20 Answered by john santu last updated on 16/Jun/20 $${y}\:=\:{wx}\:\Rightarrow\:\frac{{dy}}{{dx}}\:=\:{w}\:+\:{x}\:\frac{{dw}}{{dx}} \\ $$$$\Leftrightarrow{w}\:+\:{x}\:\frac{{dw}}{{dx}}\:=\:\frac{{wx}^{\mathrm{2}} −{w}^{\mathrm{2}} {x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{wx}^{\mathrm{2}} }…
Question Number 163934 by cortano1 last updated on 12/Jan/22 $${Let}\:{y}\left({x}\right)\:{be}\:{the}\:{solution}\:{of}\: \\ $$$$\:\:{x}^{\mathrm{2}} \:{y}''\left({x}\right)−\mathrm{2}{y}\left({x}\right)=\mathrm{0}\:\rightarrow\begin{cases}{{y}\left(\mathrm{1}\right)=\mathrm{1}}\\{{y}\left(\mathrm{2}\right)=\mathrm{1}}\end{cases} \\ $$$$\:{y}\left(\mathrm{3}\right)=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 98367 by bobhans last updated on 13/Jun/20 $$\mathrm{show}\:\mathrm{that}\:\varphi\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{2}} −\mathrm{x}^{−\mathrm{1}} \:\mathrm{is}\:\mathrm{an}\:\mathrm{explicit}\: \\ $$$$\mathrm{solution}\:\mathrm{to}\:\mathrm{linear}\:\mathrm{equation}\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:−\:\frac{\mathrm{2y}}{\mathrm{x}^{\mathrm{2}} }\:=\:\mathrm{0} \\ $$ Commented by john santu last updated…
Question Number 98255 by john santu last updated on 12/Jun/20 $$\mathrm{solve}\:\mathrm{the}\:\mathrm{following}\:\mathrm{initial}\:\mathrm{values} \\ $$$$\mathrm{DEs}\:\mathrm{20y}''\:+\:\mathrm{4y}'\:+\mathrm{y}\:=\:\mathrm{0} \\ $$$$;\:\mathrm{y}\left(\mathrm{0}\right)\:=\:\mathrm{3}.\mathrm{2}\:\mathrm{and}\:\mathrm{y}'\left(\mathrm{0}\right)\:=\:\mathrm{0}\: \\ $$ Commented by bemath last updated on 13/Jun/20 $$\mathrm{auxilary}\:\mathrm{equation}\:\mathrm{euler}−\mathrm{cauchy}…
Question Number 98106 by bobhans last updated on 11/Jun/20 $$\mathrm{y}''+\mathrm{y}\:=\:\mathrm{cos}\:\mathrm{3x}−\mathrm{2sin}\:\mathrm{3x} \\ $$ Answered by bemath last updated on 11/Jun/20 $$\mathrm{homogenous}\:\mathrm{solution}\: \\ $$$$\lambda^{\mathrm{2}} +\mathrm{1}=\mathrm{0}\:\Rightarrow\lambda\:=\:\pm\:{i} \\ $$$$\mathrm{y}_{\mathrm{h}}…
Question Number 163591 by Ar Brandon last updated on 08/Jan/22 $$\mathrm{R}\acute {\mathrm{e}soudre}\:\:\:\:\:\:\frac{\partial^{\mathrm{2}} {u}}{\partial{x}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {u}}{\partial{y}^{\mathrm{2}} }=\mathrm{10}{e}^{\mathrm{2}{x}+{y}} \\ $$ Commented by mkam last updated on 08/Jan/22…
Question Number 97922 by bemath last updated on 10/Jun/20 $$\mathrm{y}''\:+\:\mathrm{y}\:=\:\mathrm{cot}\:{x}\: \\ $$ Answered by bobhans last updated on 10/Jun/20 Commented by bemath last updated on…
Question Number 97868 by bemath last updated on 10/Jun/20 $$\mathrm{3y}'\:=\:\mathrm{2x}+\mathrm{y}−\mathrm{1}\: \\ $$ Answered by bobhans last updated on 10/Jun/20 Commented by bemath last updated on…