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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)} \\ $$$$\frac{{x}}{{ln}\left({x}\right)}=\frac{{y}\:\frac{{dy}}{{dx}}}{{ln}\left({y}\right)} \\ $$$$\frac{{xdx}}{{ln}\left({x}\right)}=\frac{{ydy}}{{ln}\left({y}\right)} \\…
Question Number 99923 by bemath last updated on 24/Jun/20 $$\mathrm{Eliminate}\:\mathrm{arbitrary}\:\mathrm{constant}\: \\ $$$${a}\:\mathrm{and}\:{b}\:\mathrm{from}\:\mathrm{z}\:=\:\left(\mathrm{x}−{a}\right)^{\mathrm{2}} +\left(\mathrm{y}−{b}\right)^{\mathrm{2}} \\ $$$$\mathrm{to}\:\mathrm{form}\:\mathrm{the}\:\mathrm{partial}\:\mathrm{differential} \\ $$$$\mathrm{equation}.\: \\ $$ Commented by bobhans last updated on…
Question Number 34051 by adil last updated on 29/Apr/18 $$\mathrm{2}{dy}/{dx}+{y}=\mathrm{0}\:\:{y}\left(\mathrm{0}\right)=−\mathrm{3} \\ $$ Answered by candre last updated on 30/Apr/18 $$\mathrm{2}\frac{{dy}}{{dx}}+{y}=\mathrm{0} \\ $$$$\frac{{dy}}{{dx}}=−\frac{{y}}{\mathrm{2}} \\ $$$$\frac{{dy}}{{y}}=−\frac{{dx}}{\mathrm{2}} \\…
Question Number 99286 by bobhans last updated on 20/Jun/20 Commented by bemath last updated on 20/Jun/20 $$\mathrm{Bernoulli}\:\mathrm{equation}\: \\ $$ Commented by bemath last updated on…
Question Number 99123 by Mr.D.N. last updated on 18/Jun/20 Commented by Rasheed.Sindhi last updated on 18/Jun/20 $${It}\:{seems}\:{that}\:{you}'{re}\:{experimenting} \\ $$$${about}\:{colour}\:{schemes}.\:{If}\:{you} \\ $$$${allow}\:{me}\:{to}\:{express}\:{my}\:{personal} \\ $$$${openion}\:{then}\:{I}\:{say}\:{that}\:{red}\:{text} \\ $$$${on}\:{black}\:{background}\:{is}\:{not}\:{so}\:…