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} \\ $$$${about}\:{colour}\:{schemes}.\:{If}\:{you} \\ $$$${allow}\:{me}\:{to}\:{express}\:{my}\:{personal} \\ $$$${openion}\:{then}\:{I}\:{say}\:{that}\:{red}\:{text} \\ $$$${on}\:{black}\:{background}\:{is}\:{not}\:{so}\:…
Question Number 99077 by Mr.D.N. last updated on 18/Jun/20 Commented by MJS last updated on 18/Jun/20 $$\mathrm{the}\:\mathrm{colour}\:\mathrm{is}\:\mathrm{nice}\:\mathrm{but}\:\mathrm{hard}\:\mathrm{to}\:\mathrm{read}… \\ $$ Commented by Rasheed.Sindhi last updated on…
Question Number 99072 by Mr.D.N. last updated on 18/Jun/20 Answered by mr W last updated on 18/Jun/20 $$\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…