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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
![(dv/dt)+kv=−bt let v=p(t)e^(−kt) ⇒(dv/dt)=(dp/dt)×e^(−kt) +p(x)×(−ke^(−kt) ) =(dp/dt)×e^(−kt) −kp(x)×e^(−kt) =(dp/dt)×e^(−kt) −kv ⇒(dp/dt)×e^(−kt) −kv+kv=−bt ⇒(dp/dt)×e^(−kt) =−bt ⇒(dp/dt)=−bte^(kt) ⇒p=−b∫te^(kt) dt=−(b/k)∫tde^(kt) =−(b/k)[te^(kt) −∫e^(kt) dt] =−(b/k)[te^(kt) −(1/k)∫e^(kt) d(kt)] =−(b/k)[te^(kt) −(1/k)e^(kt) +C] v=−(b/k)[te^(kt) −(1/k)e^(kt) +C]e^(−kt) at t=0: v=u u=−(b/k)[−(1/k)+C] ⇒C=(1/k)−((ku)/b) ⇒v=−(b/k)[te^(kt) −(1/k)e^(kt) +(1/k)−((ku)/b)]e^(−kt) ⇒v=−(b/k)[(t−(1/k))e^(kt) +(1/k)−((ku)/b)]e^(−kt) ⇒v=(b/k)[(1/k)−t+(((ku)/b)−(1/k))e^(−kt) ]](Q16548.png)
$$\frac{\mathrm{dv}}{\mathrm{dt}}+\mathrm{kv}=−\mathrm{bt} \\ $$$$\mathrm{let}\:\mathrm{v}=\mathrm{p}\left(\mathrm{t}\right)\mathrm{e}^{−\mathrm{kt}} \\ $$$$\Rightarrow\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{dp}}{\mathrm{dt}}×\mathrm{e}^{−\mathrm{kt}} +\mathrm{p}\left(\mathrm{x}\right)×\left(−\mathrm{ke}^{−\mathrm{kt}} \right) \\ $$$$=\frac{\mathrm{dp}}{\mathrm{dt}}×\mathrm{e}^{−\mathrm{kt}} −\mathrm{kp}\left(\mathrm{x}\right)×\mathrm{e}^{−\mathrm{kt}} \\ $$$$=\frac{\mathrm{dp}}{\mathrm{dt}}×\mathrm{e}^{−\mathrm{kt}} −\mathrm{kv} \\ $$$$\Rightarrow\frac{\mathrm{dp}}{\mathrm{dt}}×\mathrm{e}^{−\mathrm{kt}} −\mathrm{kv}+\mathrm{kv}=−\mathrm{bt} \\ $$$$\Rightarrow\frac{\mathrm{dp}}{\mathrm{dt}}×\mathrm{e}^{−\mathrm{kt}} =−\mathrm{bt} \\ $$$$\Rightarrow\frac{\mathrm{dp}}{\mathrm{dt}}=−\mathrm{bte}^{\mathrm{kt}} \\ $$$$\Rightarrow\mathrm{p}=−\mathrm{b}\int\mathrm{te}^{\mathrm{kt}} \mathrm{dt}=−\frac{\mathrm{b}}{\mathrm{k}}\int\mathrm{tde}^{\mathrm{kt}} \\ $$$$=−\frac{\mathrm{b}}{\mathrm{k}}\left[\mathrm{te}^{\mathrm{kt}} −\int\mathrm{e}^{\mathrm{kt}} \mathrm{dt}\right] \\ $$$$=−\frac{\mathrm{b}}{\mathrm{k}}\left[\mathrm{te}^{\mathrm{kt}} −\frac{\mathrm{1}}{\mathrm{k}}\int\mathrm{e}^{\mathrm{kt}} \mathrm{d}\left(\mathrm{kt}\right)\right] \\ $$$$=−\frac{\mathrm{b}}{\mathrm{k}}\left[\mathrm{te}^{\mathrm{kt}} −\frac{\mathrm{1}}{\mathrm{k}}\mathrm{e}^{\mathrm{kt}} +\mathrm{C}\right] \\ $$$$\mathrm{v}=−\frac{\mathrm{b}}{\mathrm{k}}\left[\mathrm{te}^{\mathrm{kt}} −\frac{\mathrm{1}}{\mathrm{k}}\mathrm{e}^{\mathrm{kt}} +\mathrm{C}\right]\mathrm{e}^{−\mathrm{kt}} \\ $$$$ \\ $$$$\mathrm{at}\:\mathrm{t}=\mathrm{0}:\:\mathrm{v}=\mathrm{u} \\ $$$$\mathrm{u}=−\frac{\mathrm{b}}{\mathrm{k}}\left[−\frac{\mathrm{1}}{\mathrm{k}}+\mathrm{C}\right] \\ $$$$\Rightarrow\mathrm{C}=\frac{\mathrm{1}}{\mathrm{k}}−\frac{\mathrm{ku}}{\mathrm{b}} \\ $$$$\Rightarrow\mathrm{v}=−\frac{\mathrm{b}}{\mathrm{k}}\left[\mathrm{te}^{\mathrm{kt}} −\frac{\mathrm{1}}{\mathrm{k}}\mathrm{e}^{\mathrm{kt}} +\frac{\mathrm{1}}{\mathrm{k}}−\frac{\mathrm{ku}}{\mathrm{b}}\right]\mathrm{e}^{−\mathrm{kt}} \\ $$$$\Rightarrow\mathrm{v}=−\frac{\mathrm{b}}{\mathrm{k}}\left[\left(\mathrm{t}−\frac{\mathrm{1}}{\mathrm{k}}\right)\mathrm{e}^{\mathrm{kt}} +\frac{\mathrm{1}}{\mathrm{k}}−\frac{\mathrm{ku}}{\mathrm{b}}\right]\mathrm{e}^{−\mathrm{kt}} \\ $$$$\Rightarrow\mathrm{v}=\frac{\mathrm{b}}{\mathrm{k}}\left[\frac{\mathrm{1}}{\mathrm{k}}−\mathrm{t}+\left(\frac{\mathrm{ku}}{\mathrm{b}}−\frac{\mathrm{1}}{\mathrm{k}}\right)\mathrm{e}^{−\mathrm{kt}} \right] \\ $$
Commented by Sai dadon. last updated on 23/Jun/17
$${Thank}\:{you}\:{Mr}. \\ $$$${Be}\:{blessed} \\ $$
Answered by ajfour last updated on 23/Jun/17
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{dv}}{{dt}}=−\left({kv}+{bt}\right)\:=−{z}\:\left({let}\right) \\ $$$$\:\:\:\frac{{dz}}{{dt}}={k}\frac{{dv}}{{dt}}+{b}\:\:\:{and}\:\:{z}={ku}\:{at}\:{t}=\mathrm{0} \\ $$$$\:{comparing}\:\boldsymbol{{k}}\frac{\boldsymbol{{dv}}}{\boldsymbol{{dt}}}=−{kz}\:=\:\frac{{dz}}{{dt}}−{b} \\ $$$$\:\Rightarrow\:\:\frac{{dz}}{{dt}}={b}−{kz} \\ $$$$\:\:\:\int_{\:{ku}} ^{\:\:{z}} \frac{{dz}}{{b}−{kz}}\:=\:\int_{\mathrm{0}} ^{\:\:{t}} {dt} \\ $$$$\:\:\:\mathrm{ln}\:\mid\frac{{b}−{kz}}{{b}−{k}^{\mathrm{2}} {u}}\mid=−{kt} \\ $$$$\:\:\:{b}−{kz}\:=\:\left({b}−{k}^{\mathrm{2}} {u}\right){e}^{−{kt}} \\ $$$$\:\:\:{b}−{k}\left({kv}+{bt}\right)\:=\:\left({b}−{k}^{\mathrm{2}} {u}\right){e}^{−{kt}} \\ $$$$\:\:\:{k}^{\mathrm{2}} {v}\:=\:{b}−{kbt}−\left({b}−{k}^{\mathrm{2}} {u}\right){e}^{−{kt}} \\ $$$$\:\:\:\:\boldsymbol{{v}}=\:\frac{\boldsymbol{{b}}}{\boldsymbol{{k}}^{\mathrm{2}} }−\frac{\boldsymbol{{b}}}{\boldsymbol{{k}}}\boldsymbol{{t}}−\left(\frac{\boldsymbol{{b}}}{\boldsymbol{{k}}^{\mathrm{2}} }−\boldsymbol{{u}}\right)\boldsymbol{{e}}^{−\boldsymbol{{kt}}} \:. \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$
Commented by Sai dadon. last updated on 23/Jun/17
$${Thanks}\:{aj}\mathrm{4} \\ $$