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Question Number 6415 by sanusihammed last updated on 26/Jun/16
y^′ (t) = C + ((v(t)y(t))/(1 + v(t)t)) Here, t is the variable y(t) is the function y^′ (t) is its first derivative v(t) is an another function in the same variable t C is a constant. Find an arbitrary solution (1) you can either find any pair of t and y(t) as a solution to this (2) you can take any value for the function v(t) (2a) v(t) can be a constant function (2b) v(t) can be a negative increasingly varying function.
$${y}^{'} \left({t}\right)\:=\:{C}\:+\:\frac{{v}\left({t}\right){y}\left({t}\right)}{\mathrm{1}\:+\:{v}\left({t}\right){t}} \\ $$$$ \\ $$$${Here},\:{t}\:\:{is}\:{the}\:{variable}\: \\ $$$${y}\left({t}\right)\:{is}\:{the}\:{function} \\ $$$${y}^{'} \left({t}\right)\:{is}\:{its}\:{first}\:{derivative}\: \\ $$$${v}\left({t}\right)\:{is}\:{an}\:{another}\:{function}\:{in}\:{the}\:{same}\:{variable}\:\:{t}\: \\ $$$${C}\:\:{is}\:{a}\:{constant}. \\ $$$$ \\ $$$${Find}\:{an}\:{arbitrary}\:{solution} \\ $$$$\left(\mathrm{1}\right)\:\:{you}\:{can}\:{either}\:{find}\:{any}\:{pair}\:{of}\:\:\:{t}\:\:{and}\:\:{y}\left({t}\right)\:\:{as}\:{a}\:{solution}\:{to} \\ $$$${this} \\ $$$$\left(\mathrm{2}\right)\:{you}\:{can}\:{take}\:{any}\:{value}\:{for}\:{the}\:{function}\:\:\:{v}\left({t}\right) \\ $$$$\left(\mathrm{2}{a}\right)\:\:{v}\left({t}\right)\:\:{can}\:{be}\:{a}\:{constant}\:{function} \\ $$$$\left(\mathrm{2}{b}\right)\:\:{v}\left({t}\right)\:\:{can}\:{be}\:{a}\:{negative}\:{increasingly}\:{varying}\:{function}. \\ $$
Answered by Chaeris27 last updated on 26/Jun/16
You should first put y(t) and y′(t) on the same side y′(t) = C + ((v(t)y(t))/(1+v(t)t)) y(t)(D−I((v(t))/(1+v(t)t))) = C Let u(t) = ((v(t))/(1+v(t)t)), then we have y′(t)−u(t)y(t) = C We can find a particular solution for C=0 y′(t) = u(t)y(t) ((y′(t))/(y(t))) = u(t) (d/dt) ln(y(t)) = u(t) ln(y(t)) = ∫u(t)dt y(t) = e^(∫u(t)dt) substituting u(t) back we have y(t) = e^(∫((v(t))/(1+v(t)t))dt) To end just add C back to get general soljtion
$${You}\:{should}\:{first}\:{put}\:{y}\left({t}\right)\:{and}\:{y}'\left({t}\right)\:{on}\:{the}\:{same}\:{side} \\ $$$${y}'\left({t}\right)\:=\:{C}\:+\:\frac{{v}\left({t}\right){y}\left({t}\right)}{\mathrm{1}+{v}\left({t}\right){t}} \\ $$$${y}\left({t}\right)\left({D}−{I}\frac{{v}\left({t}\right)}{\mathrm{1}+{v}\left({t}\right){t}}\right)\:=\:{C} \\ $$$${Let}\:{u}\left({t}\right)\:=\:\frac{{v}\left({t}\right)}{\mathrm{1}+{v}\left({t}\right){t}},\:{then}\:{we}\:{have} \\ $$$${y}'\left({t}\right)−{u}\left({t}\right){y}\left({t}\right)\:=\:{C} \\ $$$${We}\:{can}\:{find}\:{a}\:{particular}\:{solution}\:{for}\:{C}=\mathrm{0} \\ $$$${y}'\left({t}\right)\:=\:{u}\left({t}\right){y}\left({t}\right) \\ $$$$\frac{{y}'\left({t}\right)}{{y}\left({t}\right)}\:=\:{u}\left({t}\right) \\ $$$$\frac{{d}}{{dt}}\:{ln}\left({y}\left({t}\right)\right)\:=\:{u}\left({t}\right) \\ $$$${ln}\left({y}\left({t}\right)\right)\:=\:\int{u}\left({t}\right){dt} \\ $$$${y}\left({t}\right)\:=\:{e}^{\int{u}\left({t}\right){dt}} \\ $$$${substituting}\:{u}\left({t}\right)\:{back}\:{we}\:{have} \\ $$$${y}\left({t}\right)\:=\:{e}^{\int\frac{{v}\left({t}\right)}{\mathrm{1}+{v}\left({t}\right){t}}{dt}} \\ $$$${To}\:{end}\:{just}\:{add}\:{C}\:{back}\:{to}\:{get}\:{general}\:{soljtion} \\ $$
Commented by sanusihammed last updated on 26/Jun/16
Thanks so much
$${Thanks}\:{so}\:{much} \\ $$
Commented by Chaeris27 last updated on 26/Jun/16
I have checked and my solution is correct but adding C back is hard : Firstly, the general solution for C=0 is : y(t) = c_1 ∙e^(∫((v(t))/(1+v(t)t))dt) It appears that if you try to cancel out the variable part of the equation to get to simpler problem you have e^(−∫((v(t))/(1+v(t)t))dt) y(t) = c_1 e^(−∫((v(t))/(1+v(t)t))dt) y′(t) = ((v(t))/(1+v(t)t))c_1 so if you multiply the original equation by e^(−∫((v(t))/(1+v(t)t))dt) you get e^(−∫((v(t))/(1+v(t)t))dt) y′(t) − e^(−∫((v(t))/(1+v(t)t))dt) ((v(t))/(1+v(t)t))y(t) = Ce^(−∫((v(t))/(1+v(t)t))dt) Or with u(t) = ((v(t))/(1+v(t)t)) e^(−∫u(t)dt) y′(t) + (−u(t)e^(−∫u(t)dt) y(t)) = e^(−∫u(t)dt) C Let U(t) = e^(−∫u(t)dt) then U′(t) = −u(t)e^(−∫u(t)dt) which we can substitute in our equation U(t)y′(t)+U′(t)y(t) = (d/dt)(U(t)y(t)) (d/dt)(U(t)y(t)) = e^(−∫u(t)dt) C U(t)y(t) = ∫e^(−∫u(t)dt) C dt + c_1 = C∙∫e^(−∫u(t)dt) dt + c_1 (c_1 comes from the fact that we have just integrated, it can be any number) Finally divide by U(t) to get y(t) = (1/(U(t)))(C∙∫e^(−∫u(t)dt) dt + c_1 ) Substitute u(t) and U(t) (note that (1/(U(t))) = (1/e^(−∫u(t)dt) ) = e^(∫u(t)dt) ) y(t) = e^(∫((v(t))/(1+v(t)t))dt) ∙(C∙∫e^(−∫((v(t))/(1+v(t)t))dt) dt + c_1 ) So you have right here the most general solution available
$${I}\:{have}\:{checked}\:{and}\:{my}\:{solution}\:{is} \\ $$$${correct}\:{but}\:{adding}\:{C}\:{back}\:{is}\:{hard}\:: \\ $$$$ \\ $$$${Firstly},\:{the}\:{general}\:{solution}\:{for}\:{C}=\mathrm{0} \\ $$$${is}\:: \\ $$$${y}\left({t}\right)\:=\:{c}_{\mathrm{1}} \centerdot{e}^{\int\frac{{v}\left({t}\right)}{\mathrm{1}+{v}\left({t}\right){t}}{dt}} \\ $$$${It}\:{appears}\:{that}\:{if}\:{you}\:{try}\:{to}\:{cancel}\:{out} \\ $$$${the}\:{variable}\:{part}\:{of}\:{the}\:{equation}\:{to}\:{get} \\ $$$${to}\:\:{simpler}\:{problem}\:{you}\:{have} \\ $$$${e}^{−\int\frac{{v}\left({t}\right)}{\mathrm{1}+{v}\left({t}\right){t}}{dt}} \:{y}\left({t}\right)\:=\:{c}_{\mathrm{1}} \\ $$$${e}^{−\int\frac{{v}\left({t}\right)}{\mathrm{1}+{v}\left({t}\right){t}}{dt}} {y}'\left({t}\right)\:=\:\frac{{v}\left({t}\right)}{\mathrm{1}+{v}\left({t}\right){t}}{c}_{\mathrm{1}} \\ $$$${so}\:{if}\:{you}\:{multiply}\:{the}\:{original}\:{equation} \\ $$$${by}\:{e}^{−\int\frac{{v}\left({t}\right)}{\mathrm{1}+{v}\left({t}\right){t}}{dt}} \:{you}\:{get} \\ $$$${e}^{−\int\frac{{v}\left({t}\right)}{\mathrm{1}+{v}\left({t}\right){t}}{dt}} {y}'\left({t}\right)\:−\:{e}^{−\int\frac{{v}\left({t}\right)}{\mathrm{1}+{v}\left({t}\right){t}}{dt}} \frac{{v}\left({t}\right)}{\mathrm{1}+{v}\left({t}\right){t}}{y}\left({t}\right)\:=\:{Ce}^{−\int\frac{{v}\left({t}\right)}{\mathrm{1}+{v}\left({t}\right){t}}{dt}} \\ $$$${Or}\:{with}\:{u}\left({t}\right)\:=\:\frac{{v}\left({t}\right)}{\mathrm{1}+{v}\left({t}\right){t}} \\ $$$${e}^{−\int{u}\left({t}\right){dt}} {y}'\left({t}\right)\:+\:\left(−{u}\left({t}\right){e}^{−\int{u}\left({t}\right){dt}} \:{y}\left({t}\right)\right)\:=\:{e}^{−\int{u}\left({t}\right){dt}} {C} \\ $$$${Let}\:{U}\left({t}\right)\:=\:{e}^{−\int{u}\left({t}\right){dt}} \:{then}\:{U}'\left({t}\right)\:=\:−{u}\left({t}\right){e}^{−\int{u}\left({t}\right){dt}} \\ $$$${which}\:{we}\:{can}\:{substitute}\:{in}\:{our}\:{equation} \\ $$$${U}\left({t}\right){y}'\left({t}\right)+{U}'\left({t}\right){y}\left({t}\right)\:=\:\frac{{d}}{{dt}}\left({U}\left({t}\right){y}\left({t}\right)\right) \\ $$$$\frac{{d}}{{dt}}\left({U}\left({t}\right){y}\left({t}\right)\right)\:=\:{e}^{−\int{u}\left({t}\right){dt}} {C} \\ $$$${U}\left({t}\right){y}\left({t}\right)\:=\:\int{e}^{−\int{u}\left({t}\right){dt}} {C}\:{dt}\:+\:{c}_{\mathrm{1}} \\ $$$$=\:{C}\centerdot\int{e}^{−\int{u}\left({t}\right){dt}} \:{dt}\:+\:{c}_{\mathrm{1}} \\ $$$$\left({c}_{\mathrm{1}} \:{comes}\:{from}\:{the}\:{fact}\:{that}\:{we}\:{have}\right. \\ $$$$\left.{just}\:{integrated},\:{it}\:{can}\:{be}\:{any}\:{number}\right) \\ $$$${Finally}\:{divide}\:{by}\:{U}\left({t}\right)\:{to}\:{get}\: \\ $$$${y}\left({t}\right)\:=\:\frac{\mathrm{1}}{{U}\left({t}\right)}\left({C}\centerdot\int{e}^{−\int{u}\left({t}\right){dt}} \:{dt}\:+\:{c}_{\mathrm{1}} \right) \\ $$$${Substitute}\:{u}\left({t}\right)\:{and}\:{U}\left({t}\right)\:\left({note}\:{that}\right. \\ $$$$\left.\frac{\mathrm{1}}{{U}\left({t}\right)}\:=\:\frac{\mathrm{1}}{{e}^{−\int{u}\left({t}\right){dt}} }\:=\:{e}^{\int{u}\left({t}\right){dt}} \:\right) \\ $$$$ \\ $$$${y}\left({t}\right)\:=\:{e}^{\int\frac{{v}\left({t}\right)}{\mathrm{1}+{v}\left({t}\right){t}}{dt}} \centerdot\left({C}\centerdot\int{e}^{−\int\frac{{v}\left({t}\right)}{\mathrm{1}+{v}\left({t}\right){t}}{dt}} \:{dt}\:+\:{c}_{\mathrm{1}} \right) \\ $$$${So}\:{you}\:{have}\:{right}\:{here}\:{the}\:{most} \\ $$$${general}\:{solution}\:{available} \\ $$
Commented by sanusihammed last updated on 26/Jun/16
Great. i really appreciate.
$${Great}.\:{i}\:{really}\:{appreciate}. \\ $$

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