Question Number 108491 by Ar Brandon last updated on 17/Aug/20 $$\left(\mathrm{1}+\mathrm{2x}\right)\mathrm{y}''+\left(\mathrm{4x}−\mathrm{2}\right)\mathrm{y}'−\mathrm{8y}=\mathrm{0} \\ $$ Answered by Ar Brandon last updated on 20/Aug/20 $$\mathrm{6}.\:\left(\mathrm{1}+\mathrm{2x}\right)\mathrm{y}''+\left(\mathrm{4x}−\mathrm{2}\right)\mathrm{y}'−\mathrm{8y}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{2x}+\mathrm{1}\right)\left(\mathrm{y}''+\mathrm{2y}'\right)+\mathrm{4}\left(\mathrm{y}'+\mathrm{2y}\right)=\mathrm{0} \\…
Question Number 108473 by subhankar10 last updated on 17/Aug/20 $$\mathrm{3x}^{\mathrm{3}} +\mathrm{4x}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{solve}\:\mathrm{this}\:\mathrm{problem}. \\ $$ Answered by Sarah85 last updated on 17/Aug/20 $${x}^{\mathrm{3}} +\frac{\mathrm{4}}{\mathrm{3}}{x}−\mathrm{1}=\mathrm{0} \\…
Question Number 108475 by pticantor last updated on 17/Aug/20 $$\left({x}^{\mathrm{2}} +\mathrm{1}\right)\boldsymbol{{y}}^{'} +\mathrm{2}{x}\boldsymbol{{y}}={x}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$${please}\:{solve}\:{this}\:{differential} \\ $$$${equation} \\ $$ Answered by Dwaipayan Shikari last updated…
Question Number 108353 by bemath last updated on 16/Aug/20 $$\:\:\:\:\:\:\frac{\bigtriangleup\mathcal{B}{e}\mathcal{M}{ath}\bigtriangleup}{\ldots} \\ $$$$\:{General}\:{solution}\:{of}\: \\ $$$$\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\frac{{dy}}{{dx}}−\mathrm{2}{y}\:=\:\mathrm{sin}\:{x}\: \\ $$ Commented by bemath last updated on 16/Aug/20…
Question Number 108227 by Ar Brandon last updated on 15/Aug/20 $$\mathrm{y}''+\mathrm{4y}'+\mathrm{5y}=\mathrm{xe}^{−\mathrm{2x}} \mathrm{sinx} \\ $$ Answered by mathmax by abdo last updated on 15/Aug/20 $$\mathrm{h}\rightarrow\mathrm{r}^{\mathrm{2}} +\mathrm{4r}+\mathrm{5}=\mathrm{0}\rightarrow\Delta^{'}…
Question Number 173733 by Tawa11 last updated on 17/Jul/22 $$\mathrm{Show}\:\mathrm{that}\:\:\:\:\mathrm{x}^{\mathrm{3}} \:\:+\:\:\mathrm{y}^{\mathrm{3}} \:\:=\:\:\mathrm{1}\:\:\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{the}\:\mathrm{differential} \\ $$$$\mathrm{equation}\:\:\:\:\mathrm{20x}^{\mathrm{3}} \:\:\:+\:\:\:\mathrm{3y}^{\mathrm{2}} \:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:\:\:\:+\:\:\:\mathrm{6y}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{\mathrm{2}} \:\:\:=\:\:\:\:\mathrm{0} \\ $$ Commented by kaivan.ahmadi last…
Question Number 173736 by Tawa11 last updated on 17/Jul/22 $$\mathrm{Show}\:\mathrm{that}\:\:\:\:\:\mathrm{x}^{\mathrm{2}} \:\:\:+\:\:\:\mathrm{2xy}\:\:\:+\:\:\:\mathrm{3y}^{\mathrm{2}} \:\:\:=\:\:\:\mathrm{1},\:\:\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{differential}\:\mathrm{equation}\:\:\:\:\:\left(\mathrm{x}\:\:\:+\:\:\:\mathrm{3y}\right)^{\mathrm{2}} \:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:\:\:\:\:+\:\:\:\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} \:\:\:+\:\:\:\mathrm{2xy}\:\:\:+\:\:\:\mathrm{2y}^{\mathrm{3}} \right)\:\:\:=\:\:\:\mathrm{0} \\ $$ Answered by mindispower last…
Question Number 108180 by Ar Brandon last updated on 15/Aug/20 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{u}\:\mathrm{and}\:\mathrm{v}\:\mathrm{in}\:\mathrm{the}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equations}\:\mathrm{below} \\ $$$$\begin{cases}{\mathrm{u}'\left(\mathrm{e}^{−\mathrm{2x}} \mathrm{cos2x}\right)+\mathrm{v}'\left(\mathrm{e}^{−\mathrm{2x}} \mathrm{sin2x}\right)=\mathrm{0}}\\{\mathrm{u}'\left(\mathrm{e}^{−\mathrm{2x}} \mathrm{cos2x}\right)'+\mathrm{v}'\left(\mathrm{e}^{−\mathrm{2x}} \mathrm{sin2x}\right)'=\mathrm{xe}^{−\mathrm{2x}} \mathrm{sinx}}\end{cases} \\ $$$$\mathrm{where}\:\mathrm{u}\:\mathrm{and}\:\mathrm{v}\:\mathrm{are}\:\mathrm{functions}\:\mathrm{of}\:\mathrm{x} \\ $$ Terms of Service…
Question Number 108094 by Dwaipayan Shikari last updated on 14/Aug/20 $$\frac{{d}^{\mathrm{2}} \Psi}{{dt}^{\mathrm{2}} }+\frac{{d}\Psi}{{dt}}+\Psi=\mathrm{0} \\ $$ Answered by mathmax by abdo last updated on 14/Aug/20 $$\mathrm{y}^{''}…
Question Number 108053 by john santu last updated on 14/Aug/20 $$\:\:\:\:\:\:\:\:\:\:\:\frac{\curlywedge\mathcal{J}\:\mathbb{S}\curlywedge}{\:^{\rightrightarrows} } \\ $$$$\:\:\:\:\:\left(\frac{{y}}{{x}}+\sqrt{\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} }}\right){dx}={dy} \\ $$ Answered by bemath last updated on…