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

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Question Number 16546 by Sai dadon. last updated on 23/Jun/17
dv/dt=−(kv+bt) where k and b are constants solve the equation of v given v=u when t=0
dv/dt=−(kv+bt)wherekandbareconstantssolvetheequationofvgivenv=uwhent=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) ]
dvdt+kv=−btletv=p(t)e−kt⇒dvdt=dpdt×e−kt+p(x)×(−ke−kt)=dpdt×e−kt−kp(x)×e−kt=dpdt×e−kt−kv⇒dpdt×e−kt−kv+kv=−bt⇒dpdt×e−kt=−bt⇒dpdt=−btekt⇒p=−b∫tektdt=−bk∫tdekt=−bk[tekt−∫ektdt]=−bk[tekt−1k∫ektd(kt)]=−bk[tekt−1kekt+C]v=−bk[tekt−1kekt+C]e−ktatt=0:v=uu=−bk[−1k+C]⇒C=1k−kub⇒v=−bk[tekt−1kekt+1k−kub]e−kt⇒v=−bk[(t−1k)ekt+1k−kub]e−kt⇒v=bk[1k−t+(kub−1k)e−kt]
Commented by Sai dadon. last updated on 23/Jun/17
Thank you Mr. Be blessed
ThankyouMr.Beblessed
Answered by ajfour last updated on 23/Jun/17
(dv/dt)=−(kv+bt) =−z (let) (dz/dt)=k(dv/dt)+b and z=ku at t=0 comparing k(dv/dt)=−kz = (dz/dt)−b ⇒ (dz/dt)=b−kz ∫_( ku) ^( z) (dz/(b−kz)) = ∫_0 ^( t) dt ln ∣((b−kz)/(b−k^2 u))∣=−kt b−kz = (b−k^2 u)e^(−kt) b−k(kv+bt) = (b−k^2 u)e^(−kt) k^2 v = b−kbt−(b−k^2 u)e^(−kt) v= (b/k^2 )−(b/k)t−((b/k^2 )−u)e^(−kt) .
dvdt=−(kv+bt)=−z(let)dzdt=kdvdt+bandz=kuatt=0comparingkdvdt=−kz=dzdt−b⇒dzdt=b−kz∫kuzdzb−kz=∫0tdtln∣b−kzb−k2u∣=−ktb−kz=(b−k2u)e−ktb−k(kv+bt)=(b−k2u)e−ktk2v=b−kbt−(b−k2u)e−ktv=bk2−bkt−(bk2−u)e−kt.
Commented by Sai dadon. last updated on 23/Jun/17
Thanks aj4
Thanksaj4

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