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Category: Differential Equation

dv-dt-kv-bt-where-k-and-b-are-constants-solve-the-equation-of-v-given-v-u-when-t-0-

Question Number 16546 by Sai dadon. last updated on 23/Jun/17 $${dv}/{dt}=−\left({kv}+{bt}\right)\:{where}\:{k}\:{and}\:{b}\:{are}\:{constants} \\ $$$${solve}\:{the}\:{equation}\:{of}\:{v}\:{given} \\ $$$${v}={u}\:{when}\:{t}=\mathrm{0} \\ $$ Answered by mrW1 last updated on 23/Jun/17 $$\frac{\mathrm{dv}}{\mathrm{dt}}+\mathrm{kv}=−\mathrm{bt}…

Help-plz-Given-2d-2-x-dt-2-2dx-dt-x-ksint-where-k-is-constant-show-that-if-x-n-and-dx-dt-0-when-t-0-then-x-e-1-2t-n-2-5k-cos-t-2-n-4-5k-sin-t-2-1-5ksint-2-5kcost-

Question Number 16514 by Sai dadon. last updated on 23/Jun/17 $${Help}\:{plz} \\ $$$${Given}\:\mathrm{2}{d}^{\mathrm{2}} {x}/{dt}^{\mathrm{2}} +\mathrm{2}{dx}/{dt}+{x}={ksint} \\ $$$${where}\:{k}\:{is}\:{constant}\:{show}\:{that}\:{if}\:{x}={n} \\ $$$${and}\:{dx}/{dt}=\mathrm{0}\:{when}\:{t}=\mathrm{0}\:{then}\: \\ $$$$\left.{x}={e}^{−\mathrm{1}/\mathrm{2}{t}} \left\{\left({n}+\mathrm{2}/\mathrm{5}{k}\right){cos}\left({t}/\mathrm{2}\right)+\right)+\left({n}+\mathrm{4}/\mathrm{5}{k}\right){sin}\left({t}/\mathrm{2}\right)\right\} \\ $$$$−\mathrm{1}/\mathrm{5}{ksint}−\mathrm{2}/\mathrm{5}{kcost}. \\…

use-L-f-t-to-find-laplace-transform-of-d-2-x-dt-2-n-2-x-kcoswt-given-that-x-0-and-dx-dt-0-when-t-0-PLZ-HELP-

Question Number 16475 by Sai dadon. last updated on 22/Jun/17 $${use}\:{L}\left\{{f}'\left({t}\right)\right\}\:{to}\:{find}\:{laplace}\: \\ $$$${transform}\:{of}\:{d}^{\mathrm{2}} {x}/{dt}^{\mathrm{2}} +{n}^{\mathrm{2}} {x}={kcoswt} \\ $$$${given}\:{that}\:{x}=\mathrm{0}\:{and}\:{dx}/{dt}=\mathrm{0}\:{when}\:{t}=\mathrm{0} \\ $$$${PLZ}\:{HELP}. \\ $$$$ \\ $$ Answered…

Solve-the-PDE-by-method-of-separating-variables-2-u-x-2-2t-2-u-x-t-4u-0-

Question Number 81937 by Joel578 last updated on 16/Feb/20 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{PDE}\:\mathrm{by}\:\mathrm{method}\:\mathrm{of}\:\mathrm{separating}\:\mathrm{variables} \\ $$$$\frac{\partial^{\mathrm{2}} {u}}{\partial{x}^{\mathrm{2}} }\:+\:\mathrm{2}{t}\frac{\partial^{\mathrm{2}} {u}}{\partial{x}\partial{t}}\:−\:\mathrm{4}{u}\:=\:\mathrm{0} \\ $$ Commented by Joel578 last updated on 16/Feb/20 $$\mathrm{My}\:\mathrm{approach}…

Question-81824

Question Number 81824 by M±th+et£s last updated on 15/Feb/20 Answered by mind is power last updated on 17/Feb/20 $${let}\:{ln}\left({x}\right)={t} \\ $$$$\Rightarrow{y}\left({x}\right)={y}\left({e}^{{t}} \right)={z}\left({t}\right) \\ $$$$\frac{{dy}}{{dx}}={z}'\left({t}\right).\frac{\mathrm{1}}{{x}},\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}}…