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…
Question Number 108047 by bemath last updated on 14/Aug/20 $$\:\:\:\:\:\:\frac{°\bullet\mathcal{B}{e}\mathcal{M}{ath}\bullet°}{\Sigma} \\ $$$$\:\:\:{y}−{x}\:\frac{{dy}}{{dx}}\:=\:{a}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \:\frac{{dy}}{{dx}}\right)\: \\ $$ Answered by john santu last updated on 14/Aug/20 $$\:\:\:\:\:\:\frac{\diamondsuit\mathcal{JS}\bigstar}{\checkmark} \\…
Question Number 108022 by Ar Brandon last updated on 13/Aug/20 $$\left(\frac{\mathrm{d}^{\mathrm{2}} }{\mathrm{dx}^{\mathrm{2}} }−\mathrm{2x}\frac{\mathrm{d}}{\mathrm{dx}}+\mathrm{2n}\right)\mathrm{H}_{\mathrm{n}} \left(\mathrm{x}\right)=\mathrm{0} \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{first}-\mathrm{4}\:\mathrm{polynomials}\:\mathrm{of}\:\mathrm{Hermite}\left(\mathrm{H}_{\mathrm{n}} \right) \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 108018 by Ar Brandon last updated on 13/Aug/20 $$\mathrm{i}.\:\:\mathrm{x}^{\mathrm{3}} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} \mathrm{y}+\mathrm{2x}^{\mathrm{4}} =\mathrm{0} \\ $$$$\mathrm{ii}.\:\:\frac{\mathrm{dy}}{\mathrm{dx}}=−\mathrm{2}−\mathrm{y}+\mathrm{y}^{\mathrm{2}} \\ $$$$\mathrm{iii}.\:\:\mathrm{2cos}\left(\mathrm{x}\right)\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{2cos}^{\mathrm{2}} \left(\mathrm{x}\right)−\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{y}^{\mathrm{2}} \:;\:\mathrm{y}\left(\mathrm{0}\right)=−\mathrm{1} \\ $$ Answered…
Question Number 108005 by Ar Brandon last updated on 13/Aug/20 $$\mathrm{1}.\:\:\mathrm{x}''\left(\mathrm{t}\right)+\mathrm{x}\left(\mathrm{t}\right)=\mathrm{tcos}\left(\mathrm{2t}\right)+\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\mathrm{sin}\left(\mathrm{2t}\right) \\ $$$$\mathrm{2}.\:\:\mathrm{x}''\left(\mathrm{t}\right)+\mathrm{x}\left(\mathrm{t}\right)=\mathrm{t}^{\mathrm{2}} \mathrm{cos}\left(\mathrm{2t}\right) \\ $$ Answered by mathmax by abdo last updated on…
Question Number 107995 by Ar Brandon last updated on 13/Aug/20 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equation}; \\ $$$$\mathrm{x}''\left(\mathrm{t}\right)+\mathrm{2x}'\left(\mathrm{t}\right)+\mathrm{x}\left(\mathrm{t}\right)=\mathrm{1}+\mathrm{t} \\ $$$$\left(\mathrm{using}\:\mathrm{the}\:\mathrm{method}\:\mathrm{of}\:\mathrm{variation}\:\mathrm{of}\:\mathrm{parameters}\right) \\ $$ Answered by Ar Brandon last updated on 13/Aug/20…