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Category: Differential Equation

Solve-the-following-DE-y-dy-dx-6x-5y-0-x-0-y-0-

Question Number 1352 by 112358 last updated on 24/Jul/15 $${Solve}\:{the}\:{following}\:{DE}\:: \\ $$$${y}\frac{{dy}}{{dx}}+\mathrm{6}{x}+\mathrm{5}{y}=\mathrm{0}\:\:\:\:\:\left({x}\neq\mathrm{0},{y}\neq\mathrm{0}\right) \\ $$ Answered by imhunter last updated on 27/Jul/15 $${q}.{no}.\mathrm{1352}\:\:\:\:\:\:\:\:\:\:\:{y}\:{dy}/{dx}=−\mathrm{6}{x}−\mathrm{5}{y} \\ $$$$ \\…

find-general-solution-y-64x-y-2-

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}}…

Is-there-a-solution-of-y-in-terms-of-x-for-the-following-D-E-dy-dx-c-1-y-c-2-x-c-3-2-c-4-Here-c-1-c-2-c-3-c-4-are-constants-

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}.\: \\…

m-dv-dt-q-v-B-E-f-v-0-v-0-v-t-v-dr-dt-r-0-0-r-t-

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…

2-u-x-2-v-1-2-u-x-t-v-2-2-2-u-t-2-u-x-0-f-x-u-t-x-0-g-x-

Question Number 776 by 123456 last updated on 12/Mar/15 $$\frac{\partial^{\mathrm{2}} {u}}{\partial{x}^{\mathrm{2}} }={v}_{\mathrm{1}} \frac{\partial^{\mathrm{2}} {u}}{\partial{x}\partial{t}}+{v}_{\mathrm{2}} ^{\mathrm{2}} \frac{\partial^{\mathrm{2}} {u}}{\partial{t}^{\mathrm{2}} } \\ $$$${u}\left({x},\mathrm{0}\right)={f}\left({x}\right) \\ $$$${u}_{{t}} \left({x},\mathrm{0}\right)={g}\left({x}\right) \\ $$…

f-R-R-g-R-R-d-fg-dx-df-dx-dg-dx-d-f-2-dx-df-dx-df-dx-d-g-2-dx-

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…

obtain-the-series-solution-of-the-differential-equation-y-II-xy-I-y-x-2-1-y-0-1-and-y-I-0-2-

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…