Question Number 132398 by bramlexs22 last updated on 14/Feb/21 $$\mathrm{find}\:\mathrm{general}\:\mathrm{solution}\:\:\mathrm{y}'\:=\:\left(\mathrm{64x}+\mathrm{y}\right)^{\mathrm{2}} \: \\ $$ Answered by EDWIN88 last updated on 14/Feb/21 $$\mathrm{let}\:\mathrm{v}\:=\:\mathrm{64x}+\mathrm{y}\:\Leftrightarrow\frac{\mathrm{dv}}{\mathrm{dx}}\:=\:\mathrm{64}+\frac{\mathrm{dy}}{\mathrm{dx}} \\ $$$$\mathrm{or}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{dv}}{\mathrm{dx}}−\mathrm{64}\: \\ $$$$\:\Leftrightarrow\:\frac{\mathrm{dv}}{\mathrm{dx}}\:\:−\:\mathrm{64}\:=\:\mathrm{v}^{\mathrm{2}}…
Question Number 132346 by Study last updated on 13/Feb/21 $$\left({y}^{\mathrm{4}} −\mathrm{3}{x}^{\mathrm{2}} \right){dy}+{xydx}=\mathrm{0}\:\:\:\:{y}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 1211 by 112358 last updated on 14/Jul/15 $${Is}\:{there}\:{a}\:{solution}\:{of}\:{y}\:{in}\:{terms} \\ $$$${of}\:{x}\:{for}\:{the}\:{following}\:{D}.{E}? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\frac{{dy}}{{dx}}+\frac{{c}_{\mathrm{1}} }{{y}\left({c}_{\mathrm{2}} {x}+{c}_{\mathrm{3}} \right)^{\mathrm{2}} }={c}_{\mathrm{4}} \\ $$$${Here}\:{c}_{\mathrm{1}} ,\:{c}_{\mathrm{2}} ,\:{c}_{\mathrm{3}} ,\:{c}_{\mathrm{4}} \:{are}\:{constants}.\: \\…
Question Number 970 by 123456 last updated on 09/May/15 $${m}\frac{{d}\boldsymbol{{v}}}{{dt}}={q}\left(\boldsymbol{{v}}×\boldsymbol{{B}}+\boldsymbol{{E}}\right)+\boldsymbol{{f}} \\ $$$$\boldsymbol{{v}}\left(\mathrm{0}\right)=\boldsymbol{{v}}_{\mathrm{0}} \\ $$$$\boldsymbol{{v}}\left({t}\right)=? \\ $$$$\boldsymbol{{v}}=\frac{{d}\boldsymbol{{r}}}{{dt}} \\ $$$$\boldsymbol{{r}}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\boldsymbol{{r}}\left({t}\right)=?? \\ $$ Terms of Service…
Question Number 776 by 123456 last updated on 12/Mar/15 $$\frac{\partial^{\mathrm{2}} {u}}{\partial{x}^{\mathrm{2}} }={v}_{\mathrm{1}} \frac{\partial^{\mathrm{2}} {u}}{\partial{x}\partial{t}}+{v}_{\mathrm{2}} ^{\mathrm{2}} \frac{\partial^{\mathrm{2}} {u}}{\partial{t}^{\mathrm{2}} } \\ $$$${u}\left({x},\mathrm{0}\right)={f}\left({x}\right) \\ $$$${u}_{{t}} \left({x},\mathrm{0}\right)={g}\left({x}\right) \\ $$…
Question Number 767 by 123456 last updated on 09/Mar/15 $${f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${g}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$$\frac{{d}\left({fg}\right)}{{dx}}=\frac{{df}}{{dx}}\centerdot\frac{{dg}}{{dx}} \\ $$$$\frac{{d}\left({f}^{\mathrm{2}} \right)}{{dx}}=\frac{{df}}{{dx}}\centerdot\frac{{df}}{{dx}} \\ $$$$\frac{{d}\left({g}^{\mathrm{2}} \right)}{{dx}}=? \\ $$ Commented by 123456…
Question Number 757 by 123456 last updated on 07/Mar/15 $${k}\frac{{d}^{\mathrm{2}} {i}}{{dt}^{\mathrm{2}} }+{l}\frac{{di}}{{dt}}+{ri}={v} \\ $$$${i}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${i}'\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${k},{l},{r},{v}\:{are}\:{constants} \\ $$ Commented by prakash jain last…
Question Number 131805 by Engr_Jidda last updated on 08/Feb/21 $${obtain}\:{the}\:{series}\:{solution}\:{of}\:{the}\:{differential}\: \\ $$$${equation}:\:{y}^{{II}} +{xy}^{{I}} −{y}={x}^{\mathrm{2}} +\mathrm{1} \\ $$$${y}\left(\mathrm{0}\right)=\mathrm{1}\:{and}\:{y}^{{I}} \left(\mathrm{0}\right)=\mathrm{2} \\ $$ Answered by physicstutes last updated…
Question Number 730 by 123456 last updated on 04/Mar/15 $$\mathrm{sin}\:{t}={i}\frac{{di}}{{dt}} \\ $$$${i}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${i}\left({t}\right)=? \\ $$ Commented by malwaan last updated on 05/Mar/15 $$−{cos}\:{t}=\frac{{i}^{\mathrm{2}} }{\mathrm{2}}+{C}\:…
Question Number 131644 by Ahmed1hamouda last updated on 07/Feb/21 Commented by Ahmed1hamouda last updated on 07/Feb/21 $$ \\ $$$$\mathrm{solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equation} \\ $$ Answered by rs4089 last…