Question Number 100337 by I want to learn more last updated on 26/Jun/20 $$\mathrm{Solve}:\:\:\:\left(\mathrm{1}\:+\:\:\mathrm{x}^{\mathrm{2}} \right)\mathrm{y}''\:\:−\:\:\mathrm{4xy}'\:\:+\:\:\mathrm{6y}\:\:\:=\:\:\mathrm{0} \\ $$ Answered by mathmax by abdo last updated on…
Question Number 100323 by bobhans last updated on 26/Jun/20 $$\mathbb{S}\mathrm{olve}\:\mathrm{x}^{\mathrm{2}} \mathrm{y}''+\mathrm{2xy}'−\mathrm{2y}=\mathrm{0} \\ $$ Commented by bemath last updated on 26/Jun/20 $$\mathrm{since}\:\mathrm{diff}\:\mathrm{a}\:\mathrm{power}\:\mathrm{pushes}\:\mathrm{down}\:\mathrm{the} \\ $$$$\mathrm{exponent}\:\mathrm{by}\:\mathrm{one}\:\mathrm{unit},\:\mathrm{the}\:\mathrm{form}\:\mathrm{of} \\ $$$$\mathrm{this}\:\mathrm{eq}\:\mathrm{suggest}\:\mathrm{that}\:\mathrm{we}\:\mathrm{look}\:\mathrm{for}…
Question Number 34746 by candre last updated on 10/May/18 $${y}+\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{\mathrm{1}}{{x}}\left({x}\mathrm{ln}\:{x}−\frac{{dy}}{{dx}}+\mathrm{cos}\:{x}\right) \\ $$$${y}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 100184 by bobhans last updated on 25/Jun/20 $$\frac{\mathrm{ydx}\:+\:\mathrm{xdy}}{\mathrm{1}−\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} }\:+\:\mathrm{xdx}\:=\:\mathrm{0} \\ $$ Answered by maths mind last updated on 26/Jun/20 $${u}={xy}\Rightarrow{du}={ydx}+{xdy} \\ $$$$\Leftrightarrow\frac{{du}}{\mathrm{1}−{u}^{\mathrm{2}}…
Question Number 100186 by bemath last updated on 25/Jun/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 100065 by kungmikami last updated on 24/Jun/20 Answered by abdomathmax last updated on 24/Jun/20 $$\mathrm{B}\left(\mathrm{m},\mathrm{n}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{x}^{\mathrm{m}−\mathrm{1}} \left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{n}−\mathrm{1}} \:\mathrm{dx} \\ $$$$\mathrm{changement}\:\mathrm{1}−\mathrm{x}=\mathrm{t}\:\mathrm{give}\: \\ $$$$\mathrm{B}\left(\mathrm{m},\mathrm{n}\right)\:=−\int_{\mathrm{0}}…
Question Number 100048 by bramlex last updated on 24/Jun/20 Answered by mr W last updated on 24/Jun/20 $${u}=\frac{{dy}}{{dx}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{{du}}{{dx}}=\frac{{du}}{{dy}}×\frac{{dy}}{{dx}}={u}\frac{{du}}{{dy}} \\ $$$$\mathrm{2}{yu}\frac{{du}}{{dy}}−{u}^{\mathrm{2}} =\mathrm{1}…
Question Number 99992 by bemath last updated on 24/Jun/20 $$\mathrm{y}\left(\mathrm{1}+\mathrm{x}^{\mathrm{3}} \right)\mathrm{dy}−\mathrm{x}^{\mathrm{2}} \mathrm{dx}\:=\:\mathrm{0}\:;\:\mathrm{y}\left(\mathrm{2}\right)=\mathrm{3}\: \\ $$ Answered by john santu last updated on 24/Jun/20 $$\mathrm{y}\:\mathrm{dy}\:=\:\frac{\mathrm{x}^{\mathrm{2}} \:\mathrm{dx}}{\mathrm{1}+\mathrm{x}^{\mathrm{3}} }…
Question Number 99989 by bemath last updated on 24/Jun/20 $$\left(\mathrm{D}^{\mathrm{2}} −\mathrm{4D}+\mathrm{4}\right)\mathrm{y}\:=\:{xe}^{\mathrm{2}{x}} \\ $$ Answered by bobhans last updated on 24/Jun/20 $$\mathrm{homogenous}\:\mathrm{solution} \\ $$$$\beta^{\mathrm{2}} −\mathrm{4}\beta+\mathrm{4}=\mathrm{0}\:,\:\mathrm{which}\:\mathrm{has}\:\mathrm{root}\:\beta=\mathrm{2} \\…
Question Number 165503 by metamorfose last updated on 02/Feb/22 $${y}'\left({x}\right)=\frac{{xln}\left({y}\left({x}\right)\right)}{{ln}\left({x}\right){y}\left({x}\right)}\:,\:{y}\left({x}\right)=???. \\ $$ Answered by TheSupreme last updated on 02/Feb/22 $$\frac{{yy}'}{{ln}\left({y}\right)}=\frac{{x}}{{ln}\left({x}\right)} \\ $$$$\frac{{x}}{{ln}\left({x}\right)}=\frac{{y}\:\frac{{dy}}{{dx}}}{{ln}\left({y}\right)} \\ $$$$\frac{{xdx}}{{ln}\left({x}\right)}=\frac{{ydy}}{{ln}\left({y}\right)} \\…